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RD Sharma Class 12 Exercise 18.29 Indefinite Integrals Solutions Maths - Download PDF Free Online

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RD Sharma Class 12 Exercise 18.29 Indefinite Integrals Solutions Maths - Download PDF Free Online

RD Sharma Class 12 Exercise 18.29 Indefinite Integrals Solutions Maths - Download PDF Free Online

Updated on 24 Jan 2022, 10:24 AM IST

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RD Sharma Class 12 Solutions Chapter18 Indefinite Integrals - Other Exercise

Indefinite Integrals Excercise:18.29

Indefinite Integrals Exercise 18.29 Question 1

Answer : $-\frac{1}{3}\left(x^{2}-x+1\right)^{\frac{3}{2}}+\frac{3}{8}\left(2 x-1 \sqrt{x^{2}-x+1}+3 \ln \left|x+\frac{1}{2}+\sqrt{x^{2}-x+1}\right|+C\right.$
Hint: To solve the given integration, we express the linear term as a derivative of quadratic into constant plus another constant
Given : $\int(x+1) \sqrt{x^{2}-x+1} d x$
Solution :
$\begin{aligned} &x+1=a \frac{d}{d x}\left(x^{2}-x+1\right)+b \\ &\Rightarrow x+1=a(2 x-1)+b \\ &\Rightarrow x+1=2 a x-a+b \end{aligned}$
Comparing the coefficient of x and constant terms, we get
$\begin{aligned} &\Rightarrow 2 a=1 \Rightarrow a=\frac{1}{2} \\ &\Rightarrow b-a=1 \Rightarrow b=1+a \Rightarrow b=1+\frac{1}{2} \Rightarrow b=\frac{3}{2} \end{aligned}$
$\begin{aligned} &I=\int\left(\frac{1}{2}(2 x-1)+\frac{3}{2}\right) \sqrt{x^{2}-x+1} d x \\ &I=\int \frac{1}{2}(2 x-1) \sqrt{x^{2}-x+1} d x+\int \frac{3}{2} \sqrt{x^{2}-x+1} d x \end{aligned}$
For first integral let $x^{2}-x+1=t \Rightarrow(2 x-1) d x=d t$
$\begin{aligned} &I=\frac{1}{2} \int \sqrt{t} d t+\frac{3}{2} \int \sqrt{x^{2}-x+1} d x \\ &I=\frac{1}{2} \int \sqrt{t} d t+\frac{3}{2} \int \sqrt{x^{2}-2 x\left(\frac{1}{2}\right)+\left(\frac{1}{2}\right)^{2}+1-\left(\frac{1}{2}\right)^{2}} d x \end{aligned}$
$I=\frac{1}{2} \frac{t^{\frac{1}{2}+1}}{\frac{3}{2}}+\frac{3}{2} \int \sqrt{\left(x-\frac{1}{2}\right)^{2}+\frac{3}{4}} d x$
$I=\frac{1}{3} t^{3 / 2}+\frac{3}{2} \int \sqrt{\left(x-\frac{1}{2}\right)^{2}+\left(\frac{\sqrt{3}}{2}\right)^{2}} d x$
Usinf formula, $\int \sqrt{x^{2}+a^{2}} d x=\frac{x}{2} \sqrt{x^{2}+a^{2}}+\frac{a^{2}}{2} \log \left|x+\sqrt{x^{2}+a^{2}}\right|+C$
$\\I=\frac{1}{3} t^{3 / 2}+\frac{3}{2}\left(\frac{x-\frac{1}{2}}{2}\left(\sqrt{\left(\left(x-\frac{1}{2}\right)^{2}+\left(\frac{\sqrt{3}}{2}\right)^{2}\right)}\right)+\frac{\left(\frac{\sqrt{3}}{2}\right)^{2}}{2} \log \left|x-\frac{1}{2}+\sqrt{\left(x-\frac{1}{2}\right)^{2}+\left(\frac{\sqrt{3}}{2}\right)^{2}}\right|\right)+C$
$\begin{aligned} &I=\frac{1}{3} t^{3 / 2}+\frac{3}{2}\left(\frac{2 x-1}{4} \sqrt{x^{2}-x+1}+\frac{3}{8} \log \left|x-\frac{1}{2}+\sqrt{x^{2}-x+1}\right|\right)+C \\ &I=\frac{1}{3} t^{3 / 2}+\frac{3}{2}\left(\frac{2 x-1}{4} \sqrt{x^{2}-x+1}+\frac{3}{8} \log \left|x-\frac{1}{2}+\sqrt{x^{2}-x+1}\right|\right)+C \end{aligned}$
$\\I=\frac{1}{3}\left(x^{2}-x+1\right)^{3 / 2}+\frac{3}{8}(2 x-1) \sqrt{x^{2}-x+1}+\frac{9}{16} \log \left[x-\frac{1}{2}+\sqrt{x^{2}-x+1}\right]+c$

Indefinite Integrals Exercise 18.29 Question 2

Answer : $\frac{1}{6}\left(2 x^{2}+3\right)^{\frac{3}{2}}+\frac{x}{2} \sqrt{x^{2}-x+1}+\frac{3 \sqrt{2}}{4} \log \left|\frac{\sqrt{2 x}+\sqrt{2 x^{2}+3}}{\sqrt{3}}\right|+C$
Hint: To solve the given integration, we express the linear term as a derivative of quadratic into constant plus another constant
Given : $\int(x+1) \sqrt{2 x^{2}+3} d x$
Solution : $\text { Let } I=\int x \sqrt{2 x^{2}+3} d x+\int \sqrt{2 x^{2}+3} d x$
$\begin{aligned} &I_{1}=\int x \sqrt{2 x^{2}+3} d x \\ &\text { Let } 2 x^{2}+3=t \\ &4 x d x=d t \\ &x \cdot \mathrm{d} x=\frac{d t}{4} \end{aligned}$
$\begin{aligned} &I_{1}=\int \sqrt{t} \frac{d t}{4} \\ &=\frac{1}{4} \int t^{\frac{1}{2}} d x \end{aligned}$
$=\frac{1}{4} \frac{t^{1 / 2+1}}{1 / 2+1} \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \quad\left[\int x^{n} d x=\frac{x^{n}+1}{n+1}\right]$
$\begin{aligned} &=\frac{1}{4} \times 2 \frac{t^{\frac{3}{2}}}{3} \\ &=\frac{t^{\frac{3}{2}}}{6}=\frac{\left(2 x^{2}+3\right)^{\frac{3}{2}}}{6}+C_{1} \end{aligned}$ ....(i)
$I_{2}=\int \sqrt{2 x^{2}+3} d x$
Use the formula : $\int \sqrt{x^{2}+a^{2}} d x=\left[\frac{x}{2} \sqrt{x^{2}+a^{2}}+\frac{a^{2}}{2} \log \left|x+\sqrt{x^{2}+a^{2}}\right|+C\right]$
$\begin{aligned} &I_{2}=\int \sqrt{2\left(x^{2}+\frac{3}{2}\right)} d x \\ &I_{2}=\sqrt{2} \int \sqrt{x^{2}+\frac{3}{2}} d x \\ &I_{2}=\sqrt{2} \int \sqrt{x^{2}+\left(\frac{\sqrt{3}}{\sqrt{2}}\right)^{2}} d x \end{aligned}$
$I_{2}=\sqrt{2}\left[\frac{x}{2} \sqrt{x^{2}+\frac{3}{2}}+\frac{3}{2 \times 2} \log \left|x+\sqrt{x^{2}+\frac{3}{2}}\right|\right]+C_{2}$ ....(ii)
Adding (i) and (ii) ; $I=I_{1}+I_{2}$
$I=\int x \sqrt{2 x^{2}+3} d x+\int \sqrt{2 x^{2}+3} d x$
$\begin{aligned} &=\frac{\left(2 x^{2}+3\right)^{\frac{3}{2}}}{6}+\frac{x}{\sqrt{2}} \sqrt{x^{2}+\frac{3}{2}}+\frac{3}{2 \sqrt{2}} \log \left|x+\sqrt{x^{2}+\frac{3}{2}}\right|+C \\ &=\frac{1}{6}\left(2 x^{2}+3\right)^{\frac{3}{2}}+\frac{x}{\sqrt{2}} \frac{\sqrt{2 x^{2}+3}}{\sqrt{2}}+\frac{3}{2 \sqrt{2}} \log \left|x+\frac{\sqrt{2 x^{2}+3}}{\sqrt{2}}\right|+C \end{aligned}$
$\begin{aligned} &=\frac{1}{6}\left(2 x^{2}+3\right)^{\frac{3}{2}}+\frac{x}{2} \sqrt{2 x^{2}+3}+\frac{3}{2 \sqrt{2}} \frac{\sqrt{2}}{\sqrt{2}} \log \left|\frac{\sqrt{2} x+\sqrt{2 x^{2}+3}}{\sqrt{2}}\right|+C \\ &\int(x+1) \sqrt{2 x^{2}+3} d x=\frac{1}{6}\left(2 x^{2}+3\right)^{\frac{3}{2}}+\frac{x}{2} \sqrt{2 x^{2}+3}+\frac{3 \sqrt{2}}{4} \log \left|\frac{\sqrt{2} x+\sqrt{2 x^{2}+3}}{\sqrt{2}}\right|+c \end{aligned}$

Indefinite Integrals Exercise 18.29 Question 3

Answer :
$\\I=-\frac{2}{3}\left(2+3 x-x^{2}\right)^{3}-\left(\frac{2 x-3}{2}\right) \sqrt{2+3 x-x^{2}}-\frac{17}{4} \sin ^{-1}\left(\frac{2 x-3}{\sqrt{17}}\right)+C$
Hint: To solve the given integration, we express the linear term as a derivative of quadratic into constant plus another constant
Given : $\int(2 x-5) \sqrt{2+3 x-x^{2}} d x$
Solution :
$\begin{aligned} &2 x-5=-(5-2 x)=-(2+3-2 x)=-2-(3-2 x) \\ &I=\int(-2-(3-2 x)) \sqrt{2+3 x-x^{2}} d x \\ &I=\int-2 \sqrt{2+3 x-x^{2}} d x-\int(3-2 x) \sqrt{2+3 x-x^{2}} d x \end{aligned}$
Again, $2+3 x-x^{2}$
$\begin{aligned} &=-\left(x^{2}-3 x-2\right) \\ &=-\left(x^{2}-2 \cdot x \cdot \frac{3}{2}+\frac{9}{4}-2-\frac{9}{4}\right) \\ &=-\left(\left(x-\frac{3}{2}\right)^{2}-\frac{17}{4}\right) \\ &=\left(\left(\sqrt{\frac{17}{4}}\right)^{2}-\left(x-\frac{3}{2}\right)^{2}\right) \end{aligned}$
Now, $I=-2 \int\left[\sqrt{\left(\sqrt{\frac{17}{4}}\right)^{2}-\left(x-\frac{3}{2}\right)^{2}}\right] \mathrm{dx}-\int(3-2 x) \sqrt{2+3 x-x^{2}} d x$
For the second integral :
Let, $2+3 x-x^{2}=t^{2}$
$\Rightarrow[(3-2 x) d x=2 t d t]$
Use the formula : $\int \sqrt{a^{2}-x^{2}} d x=\frac{x}{2} \sqrt{a^{2}-x^{2}}+\frac{a^{2}}{2} \sin ^{-1} \frac{x}{a}+C$
$\begin{aligned} &I=-2\left[\frac{\left(x-\frac{3}{2}\right)}{2} \sqrt{\frac{17}{4}-\left(x-\frac{3}{2}\right)^{2}}+\frac{\frac{17}{4}}{2} \sin ^{-1} \frac{\left(x-\frac{3}{2}\right)}{\frac{17}{4}}\right]-\int 2 t^{2} d t \\ &I=-\left(\frac{2 x-3}{2}\right) \sqrt{2+3 x-x^{2}}-\frac{17}{4} \sin ^{-1}\left(\frac{2 x-3}{\sqrt{17}}\right)-\frac{2}{3}\left(2+3 x-x^{2}\right)^{3}+C \\ &I=-\frac{2}{3}\left(2+3 x-x^{2}\right)^{3}-\left(\frac{2 x-3}{2}\right) \sqrt{2+3 x-x^{2}}-\frac{17}{4} \sin ^{-1}\left(\frac{2 x-3}{\sqrt{17}}\right)+C \end{aligned}$

Indefinite Integrals Exercise 18.29 Question 4

Answer : $\\\frac{1}{3}\left(x^{2}+x+1\right)^{\frac{3}{2}}+\frac{3(2 x+1)}{8} \sqrt{x^{2}+x+1}+\frac{9}{16} \log \left|\left(x+\frac{1}{2}\right)+\sqrt{x^{2}+x+1}\right|+C$
Hint: To solve the given integration, we express the linear term as a derivative of quadratic into constant plus another constant
Given : $\int(x+2) \sqrt{x^{2}+x+1} d x$
Solution :
$\begin{aligned} &\text { Let }(x+2)=a \frac{d}{d x}\left(x^{2}+x+1\right)+b \\ &\Rightarrow x+2=a(2 x+1)+b \\ &\Rightarrow x+2=2 a x+a+b \end{aligned}$
Now comparing the coefficients of x and the constant term, we get
$\begin{aligned} &2 a=1 \Rightarrow a=\frac{1}{2} \text { and } \\ &a+b=2 \Rightarrow b=2-a=2-\frac{1}{2}=\frac{3}{2} \\ &I=\int\left(\frac{1}{2}(2 x+1)+\frac{3}{2}\right) \sqrt{x^{2}+x+1} d x \\ &I=\int \frac{1}{2}(2 x+1) \sqrt{x^{2}+x+1} d x+\int \frac{3}{2} \sqrt{x^{2}+x+1} d x \end{aligned}$
For the first integral, let $x^{2}+x+1=t \Rightarrow(2 x+1) d x=d t$
$\begin{aligned} &I=\frac{1}{2} \int \sqrt{t} d t+\frac{3}{2} \int \sqrt{x^{2}+2(x)\left(\frac{1}{2}\right)+\left(\frac{1}{2}\right)^{2}+1-\frac{1}{4}} d x \\ &I=\frac{1}{2} \frac{t^{\frac{1}{2}+1}}{\frac{3}{2}}+\frac{3}{2} \int \sqrt{\left(x+\frac{1}{2}\right)^{2}+\frac{3}{4}} d x \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \quad\left[\int x^{n} d x=\frac{x^{n}+1}{n+1}+C\right] \end{aligned}$
$I=\frac{t^{\frac{3}{2}}}{3}+\frac{3}{2} \int \sqrt{\left(x+\frac{1}{2}\right)^{2}+\left(\frac{\sqrt{3}}{2}\right)^{2}} d x$
$\left(\text { Use the formula: } \sqrt{x^{2}+a^{2}} d x=\frac{x}{2} \sqrt{x^{2}+a^{2}}+\frac{a^{2}}{2} \log \left|x+\sqrt{x^{2}+a^{2}}\right|+C\right)$
$\begin{aligned} &I=\frac{\left(x^{2}+x+1\right)^{\frac{3}{2}}}{3}+\frac{3}{2}\left[\frac{\left(x+\frac{1}{2}\right)}{2} \sqrt{x^{2}+x+1}+\frac{\left(\frac{\sqrt{3}}{2}\right)^{2}}{2} \log \left|\left(x+\frac{1}{2}\right)+\sqrt{x^{2}+x+1}\right|\right]+C \\ &I=\frac{1}{3}\left(x^{2}+x+1\right)^{3 / 2}+\frac{3(2 x+1)}{8} \sqrt{x^{2}+x+1}+\frac{9}{16} \log \left|\left(x+\frac{1}{2}\right)+\sqrt{x^{2}+x+1}\right|+C \end{aligned}$

Indefinite Integrals Exercise 18.29 Question 5

Answer: $\\\frac{4}{3}\left(x^{2}+x+1\right)^{\frac{3}{2}}+\frac{3(2 x+1)}{8} \sqrt{x^{2}+x+1}+\frac{27}{8} \log \left|\left(x+\frac{1}{2}\right)+\sqrt{x^{2}+x+1}\right|+C$
Hint: To solve the given integration, we express the linear team as a derivative of quadratic into constant plus another constant
Given: $I=\int(4 x+1) \sqrt{x^{2}-x-2} d x$
Solution: Let, $4 x+1=M \frac{d}{d x}\left(x^{2}-x-2\right)+N$
$\begin{aligned} &\Rightarrow 4 x+1=M(2 x-1)+N \\ &\Rightarrow 4 x+1=2 M x-M+N \end{aligned}$
Now comparing the coefficients of x and the constant term, we get
$\begin{aligned} &2 M=4 \Rightarrow M=2 \text { and }-M+N=1 \Rightarrow-2+N=1 \Rightarrow N=3 \\ &I=2 \int(2 x-1) \sqrt{x^{2}-x-2} d x+3 \int \sqrt{x^{2}-x-2} d x \end{aligned}$
For first integral, let, $x^{2}-x-2=t \Rightarrow(2 x-1) d x=d t$
For second integral, $x^{2}-x-2=x^{2}-2 \cdot x \cdot \frac{1}{2}+\left(\frac{1}{2}\right)^{2}-\left(\frac{1}{2}\right)^{2}-2=\left(x-\frac{1}{2}\right)^{2}-\frac{9}{4}=\left(x-\frac{1}{2}\right)^{2}-\left(\frac{3}{2}\right)^{2}$
So, the integral becomes
$I=2 \int \sqrt{t} d t+3 \int \sqrt{\left(x-\frac{1}{2}\right)^{2}-\left(\frac{3}{2}\right)^{2}} d x$
Use the formula : $\left[\int \sqrt{x^{2}-a^{2}} d x=\frac{x}{2} \sqrt{x^{2}-a^{2}}-\frac{a^{2}}{2} \log \left|x+\sqrt{x^{2}-a^{2}}\right|\right]+C$
$\begin{aligned} &I=2 \frac{t^{\frac{1}{2}+1}}{\frac{1}{2}+1}+3\left[\frac{\left(x-\frac{1}{2}\right)}{2} \sqrt{x^{2}-x-2}-\frac{\frac{9}{4}}{2} \log \left|\left(x-\frac{1}{2}\right)+\sqrt{\left(x-\frac{1}{2}\right)^{2}-\left(\frac{3}{2}\right)^{2}}\right|\right]+C \\ &I=2 \times 2 \frac{t^{\frac{3}{2}}}{3}+\frac{3}{2}\left[\left(x-\frac{1}{2}\right) \sqrt{x^{2}-x-2}-\frac{9}{4} \log \left|\left(x-\frac{1}{2}\right)+\sqrt{x^{2}-x-2}\right|\right]+C \end{aligned}$
$\\I=\frac{4}{3}\left(x^{2}-x-2\right)^{3 / 2}+\frac{3}{4}(2 x-1)\sqrt{x^{2} x-2}-\frac{27}{8}\log \left|\left(x-\frac{1}{2}\right)+\sqrt{x^{2}-x-2}\right|+C$

Indefinite Integrals Exercise 18.29 Question 6

Answer : $\\I=\frac{1}{6}\left(2 x^{2}-6 x+5\right)^{3 / 2}-\frac{1}{2}\left[\frac{2 x-3}{2} \sqrt{x^{2}-3 x+\frac{5}{2}}+\frac{11}{8} \log \left|\frac{2 x-3}{\sqrt{2}}+\sqrt{2 x^{2}-6 x+5}\right|\right]+C$
Hint: We solve this integration by qualitative derivation.
Given: $\int(x-2) \sqrt{2 x^{2}-6 x+5} \; d x$
Let, $x-2=a \frac{d}{d x}\left(2 x^{2}-6 x+5\right)+b$
$\begin{aligned} &\Rightarrow x-2=a(4 x-6)+b \\ &\Rightarrow x-2=4 a x+b-6 a \end{aligned}$
Now comparing the coefficients of x and the constant term, we get
$\begin{aligned} &4 a=1 \Rightarrow a=\frac{1}{4} \text { and } \\ &b-6 a=-2 \Rightarrow b=6\left(\frac{1}{4}\right)-2 \Rightarrow b=\left(-\frac{1}{2}\right) \\ &I=\int\left[\frac{1}{4}(4 x-6)+\left(-\frac{1}{2}\right)\right] \sqrt{2 x^{2}-6 x+5} d x \end{aligned}$
$I=\int \frac{1}{4}(4 x-6) \sqrt{2 x^{2}-6 x+5} d x+\int-\frac{1}{2} \sqrt{2 x^{2}-6 x+5} d x$
For the first integral : Let, $2 x^{2}-6 x+5=t \Rightarrow(4 x-6) d x=d t$
$I=\frac{1}{4} \int \sqrt{t} d t+\left(-\frac{1}{2}\right) \int \sqrt{(\sqrt{2} x)^{2}-2(\sqrt{2} x)\left(\frac{3}{\sqrt{2}}\right)+\left(\frac{3}{\sqrt{2}}\right)^{2}+5-\frac{9}{4}} d x$
$\begin{aligned} &I=\frac{1}{4} \frac{t^{\frac{1}{2}+1}}{3 / 2}-\frac{1}{2} \int \sqrt{\left(\sqrt{2} x-\frac{3}{\sqrt{2}}\right)^{2}+\frac{11}{4}} d x \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \quad\left[\int x^{n} d x=\frac{x^{n}+1}{n+1}+C\right] \\ &I=\frac{1}{2 \times 3} t^{3 / 2}-\frac{1}{2} \int \sqrt{\left(\sqrt{2} x-\frac{3}{\sqrt{2}}\right)^{2}+\left(\frac{\sqrt{11}}{2}\right)^{2}} d x \end{aligned}$
Use the formula : $\left[\sqrt{x^{2}+a^{2}} d x=\frac{x}{2} \sqrt{x^{2}+a^{2}}+\frac{a^{2}}{2} \log \left|x+\sqrt{x^{2}+a^{2}}\right|\right]$
$\begin{aligned} &I=\frac{1}{6}\left(2 x^{2}-6 x+5\right)^{\frac{3}{2}}-\frac{1}{2}\left[\frac{\left(\sqrt{2} x-\frac{3}{\sqrt{2}}\right)}{2} \sqrt{2 x^{2}-6 x+5}+\frac{\frac{11}{4}}{2} \log \left|\left(\sqrt{2} x-\frac{3}{\sqrt{2}}\right)+\sqrt{2 x^{2}-6 x+5}\right|\right]+C \\ &I=\frac{1}{6}\left(2 x^{2}-6 x+5\right)^{3 / 2}-\frac{1}{2}\left[\frac{2 x-3}{2} \sqrt{x^{2}-3 x+\frac{5}{2}}+\frac{11}{8} \log \left|\frac{2 x-3}{\sqrt{2}}+\sqrt{2 x^{2}-6 x+5}\right|\right]+C \end{aligned}$

Indefinite Integrals Exercise 18.29 Question 7

Answer : $\\I=\frac{\left(x^{2}+x+1\right)}{2}+\frac{1}{2}\left[\frac{2 x+1}{4} \sqrt{x^{2}+x+1}+\frac{3}{8} \log \left|\left(x+\frac{1}{2}\right)+\sqrt{x^{2}+x+1}\right|\right]+C$
Hint : To solve the given integration, we express the linear term as a derivative of quadratic into constant plus another constant
Given : $\int(x+1) \sqrt{x^{2}+x+1} d x$
Solution :
$\begin{aligned} &I=\frac{1}{2} \int(2 x+2) \sqrt{x^{2}+x+1} d x \\ &I=\frac{1}{2} \int(2 x+1) \sqrt{x^{2}+x+1} d x+\frac{1}{2} \int 1 \sqrt{x^{2}+x+1} d x \end{aligned}$
For thr first integral: Let, $x^{2}+x+1=t^{2} \Rightarrow(2 x+1) d x=2 t d t$
$\begin{aligned} &I=\frac{1}{2} \int 2 t d t+\frac{1}{2} \int \sqrt{x^{2}+2(x)\left(\frac{1}{2}\right)+\left(\frac{1}{2}\right)^{2}+1-\frac{1}{4}} d x \\ &I=\int t d t+\frac{1}{2} \int \sqrt{\left(x+\frac{1}{2}\right)^{2}+\frac{3}{4}} d x \end{aligned}$
$I=\frac{t^{1+1}}{1+1}+\frac{1}{2} \int \sqrt{\left(x+\frac{1}{2}\right)^{2}+\left(\frac{\sqrt{3}}{2}\right)^{2}} d x\; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \quad\left[\int x^{n} d x=\frac{x^{n}+1}{n+1}+C\right]$
$\begin{aligned} &I=\frac{t^{2}}{2}+\frac{1}{2} \int \sqrt{\left(x+\frac{1}{2}\right)^{2}+\left(\frac{\sqrt{3}}{2}\right)^{2}} d x \\ &\left(\text { Use the formula: } \sqrt{x^{2}+a^{2}} d x=\frac{x}{2} \sqrt{x^{2}+a^{2}}+\frac{a^{2}}{2} \log \left|x+\sqrt{x^{2}+a^{2}}\right|+C\right) \end{aligned}$
$\begin{aligned} &I=\frac{\left(x^{2}+x+1\right)}{2}+\frac{1}{2}\left[\frac{\left(x+\frac{1}{2}\right)}{2} \sqrt{x^{2}+x+1}+\frac{\left(\frac{\sqrt{3}}{2}\right)^{2}}{2} \log \left|\left(x+\frac{1}{2}\right)+\sqrt{x^{2}+x+1}\right|\right]+C \\ &I=\frac{\left(x^{2}+x+1\right)}{2}+\frac{1}{2}\left[\frac{2 x+1}{4} \sqrt{x^{2}+x+1}+\frac{3}{8} \log \left|\left(x+\frac{1}{2}\right)+\sqrt{x^{2}+x+1}\right|\right]+C \end{aligned}$

Indefinite Integrals Exercise 18.29 Question 8

Answer : $\\I=\frac{2}{3}\left(x^{2}+4 x+3\right)^{3 / 2}-\frac{(x+2)}{2} \sqrt{x^{2}+4 x+3}+\frac{1}{2} \log \left|(x+2)+\sqrt{x^{2}+4 x+3}\right|+C$
Hint:To solve the given integration, we express the linear term as a derivative of quadratic into constant plus another constant
Given : $\int(2 x+3) \sqrt{x^{2}+4 x+3} d x$
Solution :
Let, $x^{2}+4 x+3=u^{2}$
$\begin{aligned} &\quad \Rightarrow(2 x+4) d x=2 u d u \\ &I=\int(2 x+3) \sqrt{x^{2}+4 x+3} d x \\ &I=\int(2 x+4-1) \sqrt{x^{2}+4 x+3} d x \end{aligned}$
$\begin{aligned} &I=\int(2 x+4) \sqrt{x^{2}+4 x+3} d x-\int \sqrt{x^{2}+4 x+3} \mathrm{dx} \\ &I=\int u(2 u) \mathrm{du}-\int \sqrt{(x+2)^{2}-1^{2}} d x \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \: \: \: \: \: \; \; \; \; \; \; \; \; \; \; \quad\left[x^{2}+4 x+3=(x+2)^{2}-1\right] \end{aligned}$
Use the formula : $\left[\int \sqrt{x^{2}-a^{2}} d x=\frac{x}{2} \sqrt{x^{2}-a^{2}}-\frac{a^{2}}{2} \log \left|x+\sqrt{x^{2}-a^{2}}\right|\right]+C$
And $\left[\int x^{n} d x=\frac{x^{n}+1}{n+1}+C\right]$
$\begin{aligned} &I=2 \frac{u^{3}}{3}-\left[\frac{x+2}{2} \sqrt{(x+2)^{2}-1}-\frac{1}{2} \log \left|(x+2)+\sqrt{\left(x^{2}+2\right)^{2}-1}\right|\right]+C \\ &I=\frac{2}{3}\left(x^{2}+4 x+3\right)^{3 / 2}-\frac{(x+2)}{2} \sqrt{x^{2}+4 x+3}+\frac{1}{2} \log \left|(x+2)+\sqrt{x^{2}+4 x+3}\right|+C \end{aligned}$

Indefinite Integrals Exercise 18.29 Question 9

Answer : $\\I=\frac{2}{3}\left(x^{2}-4 x+3\right)^{3 / 2}-\frac{(x-2)}{2} \sqrt{x^{2}-4 x+3}-\frac{1}{2} \log \left|(x-2)+\sqrt{x^{2}-4 x+3}\right|+C$
Hint: To solve the given integration, we express the linear term as a derivative of quadratic into constant plus another constant
Given: $\int(2 x-5) \sqrt{x^{2}-4 x+3} d x$
$\begin{aligned} &I=\int\left[[(2 x-4)-1] \sqrt{x^{2}-4 x+3}\right] d x \\ &I=\int(2 x-4) \sqrt{x^{2}-4 x+3} d x-\int \sqrt{x^{2}-4 x+3} d x \end{aligned}$
For the first integral :
Let $x^{2}-4 x+3=t$
$\begin{aligned} &\Rightarrow 2 x-4=\frac{d t}{d x} \\ &\Rightarrow(2 x-4) d x=d t \end{aligned}$
$\begin{aligned} &I=\int \sqrt{t} d t-\int \sqrt{x^{2}-4 x+3} d x \\ &I=\int t^{\frac{1}{2}} d t-\int \sqrt{(x-2)^{2}-(1)^{2}} d x\; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \quad\left[x^{2}-4 x+3=(x-2)^{2}-1\right] \end{aligned}$
Use the formula : $\left[\int \sqrt{x^{2}-a^{2}} d x=\frac{x}{2} \sqrt{x^{2}-a^{2}}-\frac{a^{2}}{2} \log \left|x+\sqrt{x^{2}-a^{2}}\right|\right]+C$
And $\left[\int x^{n} d x=\frac{x^{n}+1}{n+1}+C\right]$
$\begin{aligned} &I=\frac{t^{1 / 2}+1}{\frac{1}{2}+1}-\left[\frac{x-2}{2} \sqrt{(x-2)^{2}-1}-\frac{1}{2} \log \left|(x-2)+\sqrt{(x-2)^{2}-1}\right|\right]+C \\ &I=\frac{2}{3}\left(x^{2}-4 x+3\right)^{3 / 2}-\frac{(x-2)}{2} \sqrt{x^{2}-4 x+3}-\frac{1}{2} \log \left|(x-2)+\sqrt{x^{2}-4 x+3}\right|+C \end{aligned}$

Indefinite Integrals Exercise 18.29 Question 10

Answer : $I=\frac{1}{3}\left(x^{2}+x\right)^{3 / 2}-\frac{1}{8}(2 x+1) \sqrt{x^{2}+x}+\frac{1}{16} \log \left|\left(x+\frac{1}{2}\right)+\sqrt{x^{2}+x}\right|+C$
Hint : To solve the given integration, we express the linear term as a derivative of quadratic into constant plus another constant
Given : $\int x \sqrt{x^{2}+x} d x$
Let, $x=a \frac{d}{d x}\left(x^{2}+x\right)+b$
$\begin{aligned} &\Rightarrow x=a(2 x+1)+b \\ &\Rightarrow x=2 x a+a+b \end{aligned}$
Comparing the coefficient of x and the constant terms, we get
$\begin{aligned} &\Rightarrow 2 a=1 \quad \text { and } \quad a+b=0 \\ &\Rightarrow a=1 / 2 \; \; \; \; \;\quad \Rightarrow b=-a \\ &\; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \Rightarrow b=-\frac{1}{2} \\ &I=\int\left(\frac{1}{2}(2 x+1)+\left(-\frac{1}{2}\right)\right) \sqrt{x^{2}+x} d x \end{aligned}$
$I=\frac{1}{2} \int(2 x+1) \sqrt{x^{2}+x} d x+\int\left(-\frac{1}{2}\right) \sqrt{x^{2}+x} d x$
For the first integral : Let $x^{2}+x=t \Rightarrow(2 x+1) d x=d t$
$I=\frac{1}{2} \int \sqrt{t} d t+\left(-\frac{1}{2}\right) \int \sqrt{x^{2}+2(x)\left(\frac{1}{2}\right)+\left(\frac{1}{2}\right)^{2}-\left(\frac{1}{2}\right)^{2}} d x$
Use the formula : $\left[\int \sqrt{x^{2}-a^{2}} d x=\frac{x}{2} \sqrt{x^{2}-a^{2}}-\frac{a^{2}}{2} \log \left|x+\sqrt{x^{2}-a^{2}}\right|\right]+C$
And $\left[\int x^{n} d x=\frac{x^{n}+1}{n+1}+C\right]$
$\begin{aligned} &I=\frac{1}{2} \frac{\frac{1}{2}+1}{\frac{1}{2}+1}-\frac{1}{2} \int \sqrt{\left(x+\frac{1}{2}\right)^{2}-\left(\frac{1}{2}\right)^{2}} d x \\ &I=\frac{1}{3} t^{\frac{3}{2}}-\frac{1}{2}\left[\frac{\left(x+\frac{1}{2}\right)}{2} \sqrt{\left(x+\frac{1}{2}\right)^{2}-\left(\frac{1}{2}\right)^{2}}-\frac{\frac{1}{4}}{2} \log \left|\left(x+\frac{1}{2}\right)+\sqrt{\left(x+\frac{1}{2}\right)^{2}-\left(\frac{1}{2}\right)^{2}}\right|\right]+C \end{aligned}$
$\begin{aligned} &I=\frac{1}{3}\left(x^{2}+x\right)^{3 / 2}-\frac{1}{2}\left[\frac{(2 x+1)}{4} \sqrt{x^{2}+x}-\frac{1}{8} \log \left|\left(x+\frac{1}{2}\right)+\sqrt{x^{2}+x}\right|\right]+C \\ &I=\frac{1}{3}\left(x^{2}+x\right)^{3 / 2}-\frac{1}{8}(2 x+1) \sqrt{x^{2}+x}+\frac{1}{16} \log \left|\left(x+\frac{1}{2}\right)+\sqrt{x^{2}+x}\right|+C \end{aligned}$

Indefinite Integrals Exercise 18.29 Question 11

Answer : $\\I=\frac{1}{3}(x+3 x-18)^{3 / 2}-\frac{9}{8}(2 x+3) \sqrt{x^{2}+3 x-18}+\frac{729}{16} \log \left|\left(x+\frac{3}{2}\right)+\sqrt{x^{2}+3 x-18}\right|+C$
Hint: To solve the given integration, we express the linear term as a derivative of quadratic into constant plus another constant
Given : $\int(x-3) \sqrt{x^{2}+3 x-18} d x$
Solution : $x-3=A+B \frac{d}{d x}\left(x^{2}+3 x-18\right)$
$\Rightarrow x-3=A+B(2 x+3)$
Comparing the coefficient of x and the constant terms, we get
$\Rightarrow 1=2 B \text { and } \Rightarrow-3=A+3 B$
$\Rightarrow B=1 / 2 \; \; \; \; \; \; \; \; \quad \Rightarrow-3-3 B=A$
$\begin{aligned} &\Rightarrow-3-\frac{3}{2}=A \\ &\Rightarrow A=-\frac{9}{2} \end{aligned}$
$\begin{aligned} &I=\int\left[-\frac{9}{2}+\frac{1}{2}(2 x+3)\right] \sqrt{x^{2}+3 x-18} d x \\ &I=-\frac{9}{2} \int \sqrt{x^{2}+3 x-18} d x+\frac{1}{2} \int(2 x+3) \sqrt{x^{2}+3 x-18} d x \end{aligned}$
For the second integral:
Let $x^{2}+3 x-18=t$
$\Rightarrow(2 x+3) d x=d t$
Use the formula : $\left[\int \sqrt{x^{2}-a^{2}} d x=\frac{x}{2} \sqrt{x^{2}-a^{2}}-\frac{a^{2}}{2} \log \left|x+\sqrt{x^{2}-a^{2}}\right|\right]+C$
And $\left[\int x^{n} d x=\frac{x^{n}+1}{n+1}+C\right]$
$\\I=-\frac{9}{2}\left[\frac{\left(x+\frac{3}{2}\right)}{2} \sqrt{x^{2}+3 x-18}-\frac{81}{4 \times 2} \log \left|\left(x+\frac{3}{2}\right)+\sqrt{x^{2}+3 x-18}\right|\right]+\frac{1}{2} \frac{t^{\frac{1}{2}+1}}{\frac{1}{2}+1}+C$
$\begin{aligned} &I=-\frac{9}{2}\left[\left(\frac{2 x+3}{4}\right) \sqrt{x^{2}+3 x-18}-\frac{81}{8} \log \left|\left(x+\frac{3}{2}\right)+\sqrt{x^{2}+3 x-18}\right|+\frac{1}{3} t^{\frac{3}{2}}\right]+C \\ &I=\frac{1}{3}(x+3 x-18)^{3 / 2}-\frac{9}{8}(2 x+3) \sqrt{x^{2}+3 x-18}+\frac{729}{16} \log \left|\left(x+\frac{3}{2}\right)+\sqrt{x^{2}+3 x-18}\right|+C \end{aligned}$

Indefinite Integrals Exercise 18.29 Question 12

Answer : $I=-\frac{1}{3}\left(3-4 x-x^{2}\right)^{\frac{3}{2}}+\frac{1}{2}(x+2) \sqrt{3-4 x-x^{2}}+\frac{7}{2} \sin ^{-1}\left(\frac{x+2}{\sqrt{7}}\right)+C$
Hint: To solve the given integration, we express the linear term as a derivative of quadratic into constant plus another constant
Given : $\int(x+3) \sqrt{3-4 x-x^{2}} d x$
Solution :
$\begin{aligned} &I=\int(x+2+1) \sqrt{3-4 x-x^{2}} d x \\ &I=\int(x+2) \sqrt{3-4 x-x^{2}} d x+\int \sqrt{3-4 x-x^{2}} d x \\ &\text { Let }, 3-4 x-x^{2}=u^{2} \end{aligned}$
$\begin{aligned} &\Rightarrow(-4-2 x) d x=2 u d u \\ &\Rightarrow-2(x+2) d x=2 u d u \\ &\Rightarrow(x+2) d x=-u d u \end{aligned}$
$\begin{aligned} &I=-\int u \sqrt{u^{2}} d u+\int \sqrt{-\left(x^{2}+4 x+4-4\right)+3} d x \\ &I=-\int u^{2} d u+\int \sqrt{-(x+2)^{2}+7} d x \end{aligned}$
Use the formula : $\int \sqrt{a^{2}-x^{2}} d x=\frac{x}{2} \sqrt{a^{2}-x^{2}}+\frac{a^{2}}{2} \sin ^{-1} \frac{x}{a}+C$
And $\left[\int x^{n} d x=\frac{x^{n}+1}{n+1}+C\right]$
$\begin{aligned} &I=-\frac{u^{3}}{3}+\int \sqrt{(\sqrt{7})^{2}-(x+2)^{2}} d x \\ &I=\frac{-\left(3-4 x-x^{2}\right)^{3 / 2}}{3}+\frac{x+2}{2} \sqrt{3-4 x-x^{2}}+\frac{7}{2} \sin ^{-1}\left(\frac{x+2}{\sqrt{7}}\right)+C \end{aligned}$$I=-\frac{1}{3}\left(3-4 x-x^{2}\right)^{\frac{3}{2}}+\frac{1}{2}(x+2) \sqrt{3-4 x-x^{2}}+\frac{7}{2} \sin ^{-1}\left(\frac{x+2}{\sqrt{7}}\right)+C$

Indefinite Integrals Exercise 18.29 Question 13

Answer : $I=-\frac{1}{2}\left(4-3 x-2 x^{2}\right)^{\frac{3}{2}}-\frac{5}{4}\left(\frac{2 x+3}{2 \sqrt{2}} \sqrt{4-3 x-2 x^{2}}+\frac{17}{4} \sin ^{-1} \frac{(2 x+3)}{\sqrt{17}}\right)+C$
Hint: To solve the given integration, we express the linear team as a derivative of quadratic into constant plus another constant
Given : $\int(3 x+1) \sqrt{4-3 x-2 x^{2}} d x$
Solution :
$\begin{aligned} &\text { Let, }\; \; \quad 3 x+1=a \frac{d}{d x}\left(4-3 x-2 x^{2}\right)+b \\ &\Rightarrow 3 x+1=a(-3-4 x)+b \\ &\Rightarrow 3 x+1=-4 a x+b-3 a \end{aligned}$
Comparing the coefficient of x and the constant terms, we get
$4 a=-3 \Rightarrow a=-\frac{3}{4}$ and
$\begin{aligned} &b-3 a=1 \Rightarrow b=1+3\left(-\frac{3}{4}\right) \Rightarrow b=-\frac{5}{4} \\ &I=\int\left[-\frac{3}{4}(-3-4 x)-\frac{5}{4}\right] \sqrt{4-3 x-2 x^{2}} d x \end{aligned}$
$I=\int\left(-\frac{3}{4}\right)(-3-4 x) \sqrt{4-3 x-2 x^{2}} d x+\int-\frac{5}{4} \sqrt{4-3 x-2 x^{2}} d x$
For the first integral :
$\begin{aligned} &\text { Let } 4-3 x-2 x^{2}=t \\ &\Rightarrow(-3-4 x) d x=d t \end{aligned}$
$I=-\frac{3}{4} \int \sqrt{t} d t-\frac{5}{4} \int \sqrt{4+\left(\frac{3}{\sqrt{2}}\right)^{2}-\left[(\sqrt{2} x)^{2}+2 \sqrt{2} x\left(\frac{3}{\sqrt{2}}\right)+\left(\frac{3}{\sqrt{2}}\right)^{2}\right]} d x$
Use the formula : $\int \sqrt{a^{2}-x^{2}} d x=\frac{x}{2} \sqrt{a^{2}-x^{2}}+\frac{a^{2}}{2} \sin ^{-1} \frac{x}{a}+C$
And $\left[\int x^{n} d x=\frac{x^{n}+1}{n+1}+C\right]$
$\begin{aligned} &I=-\frac{3}{4} \frac{t^{\frac{1}{2}+1}}{\frac{1}{2}+1}-\frac{5}{4} \int \sqrt{\frac{17}{2}-\left(\sqrt{2} x+\frac{3}{\sqrt{2}}\right)^{2}} d x \\ &I=-\frac{3}{4} \frac{t^{\frac{3}{2}}}{\frac{3}{2}}-\frac{5}{4} \sqrt{\left(\frac{\sqrt{17}}{\sqrt{2}}\right)^{2}-\left(\sqrt{2} x+\frac{3}{\sqrt{2}}\right)^{2}} d x \end{aligned}$
$\begin{aligned} &I=-\frac{1}{2}\left(4-3 x-2 x^{2}\right)^{\frac{3}{2}}-\frac{5}{4}\left(\frac{\sqrt{2} x+\frac{3}{\sqrt{2}}}{2} \sqrt{4-3 x-2 x^{2}}+\frac{17}{2} \frac{1}{2} \sin ^{-1} \frac{\left(\sqrt{2} x+\frac{3}{\sqrt{2}}\right)}{\frac{\sqrt{17}}{\sqrt{2}}}\right)+C \\ &I=-\frac{1}{2}\left(4-3 x-2 x^{2}\right)^{\frac{3}{2}}-\frac{5}{4}\left(\frac{2 x+3}{2 \sqrt{2}} \sqrt{4-3 x-2 x^{2}}+\frac{17}{4} \sin ^{-1} \frac{(2 x+3)}{\sqrt{17}}\right)+C \end{aligned}$

Indefinite Integrals Exercise 18.29 Question 14

Answer : $\frac{-2}{9}\left(10-4 x-3 x^{2}\right)^{\frac{3}{2}}+\frac{11}{8}(3 x+2) \sqrt{10-4 x+3 x^{2}}+\frac{187}{9 \sqrt{3}} \sin ^{-1}\left(\frac{3 x+2}{\sqrt{34}}\right)+C$
Hint: To solve the given integration, we express the linear team as a derivative of quadratic into constant plus another constant
Given: $\int(2 x+5) \sqrt{10-4 x-3 x^{2}} d x$
Solution : $\int(2 x+5) \sqrt{10-4 x-3 x^{2}} dx$
$\begin{aligned} &\text { Let, }(2 x+5)=A \frac{d}{d x}\left(10-4 x-3 x^{2}\right)+B \\ &(2 x+5)=A(-4-6 x)+B \\ &(2 x+5)=-4 A+B-6 A x \end{aligned}$
Comparing the coefficient of x and the constant terms, we get
$\begin{aligned} &6 A=-2 \Rightarrow A=-\frac{1}{3} \text { and } \\ &-4 A+B=5 \Rightarrow B=5+4 A=5+4\left(-\frac{1}{3}\right)=\frac{11}{3} \\ &I=-\frac{1}{3} \int(-4-6 x) \sqrt{10-4 x-3 x^{2}} d x \\ &I=-\frac{1}{3} \int(-6 x-4) \sqrt{10-4 x-3 x^{2}} d x \end{aligned}$
$\begin{aligned} &I=-\frac{1}{3}\left[\int(-6 x-4) \sqrt{10-4 x-3 x^{2}} d x+\int \frac{11}{3} \sqrt{10-4 x-3 x^{2}} d x\right] \\ &I=-\frac{1}{3} \int(-6 x-4) \sqrt{10-4 x-3 x^{2}} d x+\left(\frac{11}{3}\right) \int \sqrt{10-4 x-3 x^{2}} d x \end{aligned}$
For the first integral:
$\begin{aligned} &\text { Let, } 10-4 x-3 x^{2}=t \\ &\Rightarrow(-6 x-4) d x=d t \\ &I=-\frac{1}{3} \int \sqrt{t} d t+\frac{11}{3} \times \sqrt{3} \int \sqrt{\frac{10}{3}-\frac{4}{3} x-x^{2}} d x \end{aligned}$
Use the formula : $\int \sqrt{a^{2}-x^{2}} d x=\frac{x}{2} \sqrt{a^{2}-x^{2}}+\frac{a^{2}}{2} \sin ^{-1} \frac{x}{a}+C$
And $\left[\int x^{n} d x=\frac{x^{n}+1}{n+1}+C\right]$
$\begin{aligned} &I=-\frac{1}{3} \frac{t^{\frac{3}{2}}}{\frac{3}{2}}+\frac{11}{3} \sqrt{3} \int \sqrt{\left(\frac{10}{3}\right)-\left\{x^{2}+2 \frac{2}{3} x+\left(\frac{2}{3}\right)^{2}\right\}+\left(\frac{2}{3}\right)^{2}} d x \\ &I=-\frac{2}{9} t^{3 / 2}+\frac{11}{\sqrt{3}} \int \sqrt{\left(\frac{\sqrt{34}}{3}\right)^{2}-\left(x+\frac{2}{3}\right)^{2}} d x \end{aligned}$
$\begin{aligned} &I=-\frac{2}{9}\left(10-4 x-3 x^{2}\right)^{\frac{3}{2}}+\frac{11}{\sqrt{3}}\left[\frac{1}{2}\left(x+\frac{2}{3}\right) \sqrt{\left(\frac{\sqrt{34}}{3}\right)^{2}-\left(x+\frac{2}{3}\right)^{2}}+\frac{34 / 9}{2} \sin ^{-1}\left[\frac{x+\frac{2}{3}}{\frac{\sqrt{34}}{3}}\right]\right]+C \\ &I=-\frac{2}{9}\left(10-4 x-3 x^{2}\right)^{\frac{3}{2}}+\frac{11}{\sqrt{3}}\left[\frac{\left(x+\frac{2}{3}\right)\left(\frac{10}{3}-\frac{4 x}{3}-x^{2}\right)^{\frac{1}{2}}}{2}\right]+\frac{187}{9 \sqrt{3}} \sin ^{-1}\left[\frac{3 x+2}{\sqrt{34}}\right]+C \end{aligned}$$\\I=-\frac{2}{9}\left(10-4 x-3 x^{2}\right)^{\frac{3}{2}}+\frac{11}{18}(3 x+2) \sqrt{10-4 x-3 x^{2}}+\frac{187}{9 \sqrt{3}} \sin ^{-1}\left[\frac{3 x+2}{\sqrt{34}}\right]+C$

Rd Sharma class 12 chapter 18 exercise 18.29 has around 14 inquiries, including its subparts, and it consolidates themes like: -

Assessment of integrals by utilizing mathematical replacements like first need to solve the quadratic equation then proceed with the differentiation and integration
Questions based on the Formula of Integration

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