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RD Sharma Class 12 Exercise 18.10 Indefinite Integrals Solutions Maths

Indefinite Integrals Excercise:18.10

Indefinite Integrals Exercise 18.10 Question 1

Answer: $\frac{2}{7}(x+2)^{\frac{7}{2}}-\frac{8}{5}(x+2)^{\frac{5}{2}}+\frac{8}{3}(x+2)^{\frac{3}{2}}+c$
Hint: To solve this type of problem, we use substitution method. Let $x+2=t$ then solve the integral by general method.
Given: $\int x^{2} \sqrt{x+2} d x$
Solution:
Let $I=\int x^{2} \sqrt{x+2} d x$
Substitute $x+2=t \Rightarrow d x=d t$
$\begin{aligned} I &=\int(t-2)^{2} \sqrt{t} d t \qquad\qquad(\because x=t-2) \\ & \end{aligned}$
$=\int\left(t^{2}+4-4 t\right) t^{\frac{1}{2}} d t \qquad\qquad\left[\because(a-b)^{2}=a^{2}+b^{2}-2 a b\right] \\$
$=\int\left(t^{2} \cdot t^{\frac{1}{2}}+4 t^{\frac{1}{2}}-4 t . t^{\frac{1}{2}}\right) d t \\$
$=\int\left(t^{\frac{5}{2}}+4 t^{\frac{1}{2}}-4 t^{\frac{3}{2}}\right) d t \qquad\qquad\left(\because a^{m} \cdot a^{n}=a^{m+n}\right)$
$\begin{array}{r} =\int t^{\frac{5}{2}} d t+4 \int t^{\frac{1}{2}} d t-4 \int t^{\frac{3}{2}} d t \\ \end{array}$
$\quad\left[\because \int\{f(x) \pm a g(x)\} d x=\int f(x) d x \pm a \int g(x) d x\right]$
$=\frac{t^{\frac{5}{2}+1}}{\frac{5}{2}+1}+4 \frac{t^{\frac{1}{2}+1}}{\frac{1}{2}+1}-4 \frac{t^{\frac{3}{2}+1}}{\frac{3}{2}+1}+c \qquad\qquad\left[\because \int t^{n} d t=\frac{t^{n+1}}{n+1}+c\right]$
$\begin{aligned} &=\frac{t^{\frac{7}{2}}}{\frac{7}{2}}+4 \frac{t^{\frac{3}{2}}}{\frac{3}{2}}-4 \frac{t^{\frac{5}{2}}}{\frac{5}{2}}+c \\ & \end{aligned}$
$=\frac{2}{7}(t)^{\frac{7}{2}}-4 \cdot \frac{2}{5}(t)^{\frac{5}{2}}+4 \frac{2}{3}(t)^{\frac{3}{2}}+c \\$
$=\frac{2}{7}(x+2)^{\frac{7}{2}}-\frac{8}{5}(x+2)^{\frac{5}{2}}+\frac{8}{3}(x+2)^{\frac{3}{2}}+c \qquad \qquad[\because x+2=t]$

Indefinite Integrals Exercise 18.10 Question 2

Answer: $\frac{2}{5}(x-1)^{\frac{5}{2}}+2(x-1)^{\frac{1}{2}}+\frac{4}{3}(x-1)^{\frac{3}{2}}+c$
Hint: Use substitution method to solve this type of integral
Given: $\int \frac{x^{2}}{\sqrt{x-1}} d x$
Solution:
Let $I=\int \frac{x^{2}}{\sqrt{x-1}} d x$
Substitute $x-1=t \Rightarrow d x=d t$ then
$\begin{aligned} I &=\int \frac{(t+1)^{2}}{\sqrt{t}} d t \qquad(\because x=t+1) \\ & \end{aligned}$
$=\int \frac{t^{2}+1+2 t}{\sqrt{t}} d t \qquad\left[\because(a+b)^{2}=a^{2}+b^{2}+2 a b\right] \\$
$=\int\left(\frac{t^{2}}{\sqrt{t}}+\frac{1}{\sqrt{t}}+2 \frac{t}{\sqrt{t}}\right) d t \\$
$=\int\left(t^{2-\frac{1}{2}}+t^{-\frac{1}{2}}+2 t^{1-\frac{1}{2}}\right) d t$
$\begin{aligned} &=\int\left(t^{\frac{3}{2}}+t^{-\frac{1}{2}}+2 t^{\frac{1}{2}}\right) d t \\ & \end{aligned}$
$=\int t^{\frac{3}{2}} d t+\int t^{-\frac{1}{2}} d t+2 \int t^{\frac{1}{2}} d t$
$\begin{aligned} &=\frac{t^{\frac{3}{2}}{2}}{\frac{3}{2}+1}+\frac{t^{-\frac{1}{2}+1}}{-\frac{1}{2}+1}+2 \frac{t^{\frac{1}{2}+1}}{\frac{1}{2}+1}+c \qquad\left[\because f^{n} d t=\frac{t^{n+1}}{n+1}+c\right] \\ & \end{aligned}$
$=\frac{t^{\frac{5}{2}}}{\frac{5}{2}}+\frac{t^{\frac{1}{2}}}{\frac{1}{2}}+2 \frac{t^{\frac{3}{2}}}{\frac{3}{2}}+c$
$\begin{aligned} &=\frac{2}{5}(t)^{\frac{5}{2}}+2(t)^{\frac{1}{2}}+2 \cdot \frac{2}{3}(t)^{\frac{3}{2}}+c \\ & \end{aligned}$
$=\frac{2}{5}(x-1)^{\frac{5}{2}}+2(x-1)^{\frac{1}{2}}+\frac{4}{3}(x-1)^{\frac{3}{2}}+c \qquad[\because x-1=t]$

Indefinite Integrals Exercise 18.10 Question 3

Answer: $\frac{2}{135}(3 x+4)^{\frac{5}{2}}-\frac{16}{81}(3 x+4)^{\frac{3}{2}}+\frac{32}{27}(3 x+4)^{\frac{1}{2}}+c$
Hint: Use substitution method to solve this type of integral
Given: $\int \frac{x^{2}}{\sqrt{3 x+4}} d x$
Solution:
$I=\int \frac{x^{2}}{\sqrt{3 x+4}} d x$
Put $3 x+4=t \Rightarrow 3 d x=d t \Rightarrow d x=\frac{d t}{3}$ then
$\begin{aligned} &I=\int \frac{\frac{1}{3^{2}}(t-4)^{2}}{\sqrt{t}} \frac{d t}{3} \qquad\left(\because x=\frac{t-4}{3}\right) \\ & \end{aligned}$
$\Rightarrow I=\frac{1}{3 \times 3^{2}} \int \frac{t^{2}+16-8 t}{\sqrt{t}} d t \qquad\left[\because(a-b)^{2}=a^{2}+b^{2}-2 a b\right]$
$\begin{aligned} &\Rightarrow I=\frac{1}{27} \int\left\{\frac{t^{2}}{\sqrt{t}}+16 \frac{1}{\sqrt{t}}-8 \cdot \frac{t}{\sqrt{t}}\right\} d t \\ & \end{aligned}$
$\Rightarrow I=\frac{1}{27} \int\left\{t^{2-\frac{1}{2}}+16 t^{-\frac{1}{2}}-8 t^{1-\frac{1}{2}}\right\} d t$
$\begin{aligned} &\Rightarrow I=\frac{1}{27} \int\left(t^{\frac{3}{2}}-8 t^{\frac{1}{2}}+16 t^{-\frac{1}{2}}\right) d t \\ & \end{aligned}$
$\Rightarrow I=\frac{1}{27}\left[\int t^{\frac{3}{2}} d t-8 \int t^{\frac{1}{2}} d t+16 \int t^{-\frac{1}{2}} d t\right]$
$\Rightarrow I=\frac{1}{27}\left[\frac{t^{\frac{3}{2}+1}}{\frac{3}{2}+1}-8 \cdot \frac{t^{\frac{1}{2}}{2}+1} + 16 \cdot \frac{t^{-\frac{1}{2}+1}}{-\frac{1}{2}+1}+c\right] \qquad\left[\because f t^{n} d t=\frac{t^{n+1}}{n+1}+c\right]$
$\begin{aligned} &\Rightarrow I=\frac{1}{27}\left[\frac{t^{\frac{5}{2}}}{\frac{5}{2}}-8 \frac{t^{\frac{3}{2}}}{\frac{3}{2}}+16 \frac{t^{\frac{1}{2}}}{\frac{1}{2}}+c\right] \\ & \end{aligned}$
$\Rightarrow I=\frac{1}{27}\left[\frac{2}{5}(t)^{\frac{5}{2}}-8 \cdot \frac{2}{3}(t)^{\frac{3}{2}}+16.2(t)^{\frac{1}{2}}+c\right] \\$
$\Rightarrow I=\frac{2}{135}(3 x+4)^{\frac{5}{2}}-\frac{16}{81}(3 x+4)^{\frac{3}{2}}+\frac{32}{27}(3 x+4)^{\frac{1}{2}}+c$

Indefinite Integrals exercise 18.10 question 4

Answer: $2 \log |(x-1)|-\frac{1}{x-1}+c$
Hint: Use substitution method to solve this type of integral
Given: $\int \frac{2 x-1}{(x-1)^{2}} d x$
Solution:
Let $I=\int \frac{2 x-1}{(x-1)^{2}} d x$
Put $x-1=t \Rightarrow d x=d t$ then
$I=\int \frac{2(t+1)-1}{(t)^{2}} d t$
$\begin{aligned} &\Rightarrow I=\int \frac{2 t+2-1}{t^{2}} d t=\int \frac{2 t+1}{t^{2}} d t \\ & \end{aligned}$
$\Rightarrow I=\int \frac{2 t}{t^{2}}+\frac{1}{t^{2}} d t=\int\left\{2 t^{1-2}+t^{-2}\right\} d t \\$
$\Rightarrow I=\int\left(2 t^{-1}+t^{-2}\right) d t=\int\left\{\frac{2}{t}+t^{-2}\right\} d t \\$
$\Rightarrow I=2 \int \frac{1}{t} d t+\int t^{-2} d t$
$\Rightarrow I=2 \log |t|+\frac{t^{-2+1}}{-2+1}+c \qquad\left[\int \frac{1}{t} d t=\log |t|+c \& \int f^{n} d t=\frac{t^{n+1}}{n+1}+c\right]$
$\begin{aligned} &\Rightarrow I=2 \log |t|+\frac{t^{-1}}{-1}+c \\ & \end{aligned}$
$\Rightarrow I=2 \log |(x-1)|-(x-1)^{-1}+c \\$
$\therefore I=2 \log |(x-1)|-\frac{1}{x-1}+c$

Indefinite Integrals exercise 18.10 question 5

Answer: $\frac{4}{7}(x+2)^{\frac{7}{2}}-\frac{16}{5}(x+2)^{\frac{5}{2}}+\frac{22}{3}(x+2)^{\frac{3}{2}}+c$
Hint: Use substitution method to solve this type of integral
Given: $\int\left(2 x^{2}+3\right) \sqrt{x+2} d x$
Solution: Let $I=\int\left(2 x^{2}+3\right) \sqrt{x+2} d x$
Put $x+2=t \Rightarrow d x=d t$
$\begin{aligned} &I=\int\left\{2(t-2)^{2}+3\right\} \sqrt{t} d t \qquad(\because x=t-2) \\ & \end{aligned}$
$\Rightarrow I=\int\left\{2\left(t^{2}-4 t+4\right)+3\right\} \sqrt{t} d t \qquad\left[\because(a-b)^{2}=a^{2}+b^{2}-2 a b\right]$
$\begin{aligned} &\Rightarrow I=\int\left\{2 t^{2}-8 t+8+3\right\} \sqrt{t} d t \\ & \end{aligned}$
$\Rightarrow I=\int\left(2 t^{2}-8 t+11\right) \sqrt{t} d t \\$
$\Rightarrow I=\int\left(2 t^{2} \sqrt{t}-8 t \cdot \sqrt{t}+11 \sqrt{t}\right) d t$
$\begin{aligned} &\Rightarrow I=\int\left(2 t^{2+\frac{1}{2}}-8 t^{1+\frac{1}{2}}+11 t^{\frac{1}{2}}\right) d t \\ & \end{aligned}$
$\Rightarrow I=\left[2 \int t^{\frac{5}{2}} d t-8 \int t^{\frac{3}{2}} d t+11 \int t^{\frac{1}{2}} d t\right]$
$\Rightarrow I=\left[2 \cdot \frac{t^{\frac{5}{2}+1}}{\frac{5}{2}+1}-8 \cdot \frac{t^{\frac{3}{2}}{ } \frac{1}{2}+1} + 11 \cdot \frac{t^{\frac{1}{2}+1}}{\frac{1}{2}+1}+c\right] \qquad\left[\because \int t^{n} d t=\frac{t^{n+1}}{n+1}+c\right]$
$\begin{aligned} &\Rightarrow I=\left[2 \cdot \frac{t^{\frac{7}{2}}}{\frac{7}{2}}-8 \cdot \frac{t^{\frac{5}{2}}}{\frac{5}{2}}+11 \cdot \frac{t^{\frac{3}{2}}}{\frac{3}{2}}+c\right] \\ & \end{aligned}$
$\Rightarrow I=2 \cdot \frac{2}{7} t^{\frac{7}{2}}-8 \cdot \frac{2}{5} t^{\frac{5}{2}}+11 \cdot \frac{2}{3} t^{\frac{3}{2}}+c \\$
$\Rightarrow I=\frac{4}{7}(x+2)^{\frac{7}{2}}-\frac{16}{5}(x+2)^{\frac{5}{2}}+\frac{22}{3}(x+2)^{\frac{3}{2}}+c$

Indefinite Integrals exercise 18.10 question 6

Answer: $(x+1)+\log |(x+1)|+\frac{1}{x+1}+c$
Hint: Use substitution method to solve this type of integral
Given: $\int \frac{x^{2}+3 x+1}{(x+1)^{2}} d x$
Solution: let $I=\int \frac{x^{2}+3 x+1}{(x+1)^{2}} d x$
Substitute $x+1=t \Rightarrow d x=d t$ then
$\Rightarrow I=\int \frac{(t-1)^{2}+3(t-1)+1}{t^{2}} d t \qquad(\because x=t-2)$
$\begin{aligned} &\Rightarrow I=\int \frac{t^{2}-2 t+1+3 t-3+1}{t^{2}} d t \qquad\left[\because(a-b)^{2}=a^{2}+b^{2}-2 a b\right] \\ & \end{aligned}$
$\Rightarrow I=\int \frac{t^{2}+t-1}{t^{2}} d t \\$
$\Rightarrow I=\int \frac{t^{2}}{t^{2}}+\frac{t}{t^{2}}-\frac{1}{t^{2}} d t$
$\begin{aligned} &\Rightarrow I=\int\left(1+\frac{1}{t}-t^{-2}\right) d t \\ & \end{aligned}$
$\Rightarrow I=\int t^{0} d t+\int \frac{1}{t} d t-\int t^{-2} d t$
$\Rightarrow I=\frac{t^{0+1}}{0+1}+\log |t|-\frac{t^{-2+1}}{-2+1}+c \qquad\left[\int \frac{1}{t} d t=\log |t|+c \& \int t^{n} d t=\frac{t^{n+1}}{n+1}+c\right]$
$\begin{aligned} &\Rightarrow I=t+\log |t|-\frac{t^{-1}}{-1}+c \\ & \end{aligned}$
$\Rightarrow I=t+\log |t|+\frac{1}{t}+c \\$
$\therefore I=(x+1)+\log |(x+1)|+\frac{1}{x+1}+c$

Indefinite Integrals Excercise 18.10 Question 7

Answer: $-\frac{2}{5}(1-x)^{\frac{5}{2}}+\frac{4}{3}(1-x)^{\frac{3}{2}}+-2(1-x)^{\frac{1}{2}}+c$
Hint: Use substitution method to solve this type of integral
Given: $\int \frac{x^{2}}{\sqrt{1-x}} d x$
Solution: Let $I=\int \frac{x^{2}}{\sqrt{1-x}} d x$
Substitute $1-x=t \Rightarrow d x=-d t$ then
$\begin{aligned} &I=\int \frac{(1-t)^{2}}{\sqrt{t}}(-d t) \qquad(\because x=1-t) \\ & \end{aligned}$
$\Rightarrow I=-\int\left(\frac{1+t^{2}-2 t}{\sqrt{t}}\right) d t \qquad\left[\because(a-b)^{2}=a^{2}+b^{2}-2 a b\right]$
$\begin{aligned} &\Rightarrow I=-\int\left\{\frac{1}{\sqrt{t}}+\frac{t^{2}}{\sqrt{t}}-\frac{2 t}{\sqrt{t}}\right\} d t \\ & \end{aligned}$
$\Rightarrow I=-\int\left(t^{\frac{-1}{2}}+t^{2-\frac{1}{2}}-2 t^{1-\frac{1}{2}}\right) d t \\$
$\Rightarrow I=-\left[\int t^{-\frac{1}{2}} d t+\int t^{\frac{3}{2}} d t-2 \int t^{\frac{1}{2}} d t\right.$
$\Rightarrow I=-\left[\frac{t^{-\frac{1}{2}+1}}{-\frac{1}{2}+1}+\frac{t^{\frac{3}{2}+1}}{\frac{3}{2}+1}-2 \frac{t^{\frac{1}{2}+1}}{\frac{1}{2}+1}+c\right] \qquad\left[\because \int t^{n} d t=\frac{t^{n+1}}{n+1}+c\right]$
$\begin{aligned} &\Rightarrow I=-\left[\frac{t^{\frac{1}{2}}}{\frac{1}{2}}+\frac{t^{\frac{5}{2}}}{\frac{5}{2}}-2 \frac{t^{\frac{3}{2}}}{\frac{3}{2}}+c\right] \\ & \end{aligned}$
$\Rightarrow I=-\left[2 . t^{\frac{1}{2}}+\frac{2}{5} t^{\frac{5}{2}}-2 \cdot \frac{2}{3} t^{\frac{3}{2}}\right]+c \\$
$\Rightarrow I=-2(1-x)^{\frac{1}{2}}-\frac{2}{5}(1-x)^{\frac{5}{2}}+\frac{4}{3}(1-x)^{\frac{3}{2}}+c \qquad[\because 1-x=t]$
$\therefore I=-\frac{2}{5}(1-x)^{\frac{5}{2}}+\frac{4}{3}(1-x)^{\frac{3}{2}}+-2(1-x)^{\frac{1}{2}}+c$

Indefinite Integrals Excercise 18.10 Question 7

Indefinite Integrals Excercise 18.10 Question 8

Answer: $\frac{(1-x)^{25}}{25}-\frac{(1-x)^{24}}{24}+c$
Hint: Use substitution method to solve this type of integral
Given: $\int x(1-x)^{23} d x$
Solution: let $I=\int x(1-x)^{23} d x$
Substitute $1-x=t \Rightarrow-d x=d t$ then
$\begin{aligned} &I=\int(1-t) \cdot t^{23}(-d t) \qquad(\because x=1-t) \\ & \end{aligned}$
$\Rightarrow I=-\int(1-t) \cdot t^{23} d t=-\int t^{23}-t . t^{23} d t \\$
$\Rightarrow I=-\int\left(t^{23}-t^{24}\right) d t=-\int t^{23} d t+\int t^{24} d t$
$\begin{aligned} &\Rightarrow I=-\frac{t^{23+1}}{23+1}+\frac{t^{24+1}}{24+1}+c \qquad\left[\because \int t^{n} d t=\frac{t^{n+1}}{n+1}+c\right] \\ & \end{aligned}$
$\Rightarrow I=\frac{t^{25}}{25}-\frac{t^{24}}{24}+c \\$
$\therefore I=\frac{(1-x)^{25}}{25}-\frac{(1-x)^{24}}{24}+c$

Indefinite Integrals Excercise 18.10 Question 9

Answer: $2 \sqrt{x}-4 x^{\frac{1}{4}}+4 \log \left|1+x^{\frac{1}{4}}\right|+c$
Hint: Use substitution method to solve this type of integral
Given: $\int \frac{1}{\sqrt{x}+\sqrt[4]{x}} d x$
Solution:
Let $I=\int \frac{1}{\sqrt{x}+\sqrt[4]{x}} d x$
In the given integral, the exponent of $x \text { are } \frac{1}{2} \& \frac{1}{4}$ and the LCM of denominators is 4
Substitute $x=t^{4} \Rightarrow d x=4 t^{3} d t$ then
$\begin{aligned} &\Rightarrow I=\int \frac{1}{t^{2}+t}\left(4 t^{3}\right) d t \qquad\left(\because x=t^{4}\right) \\ & \end{aligned}$
$\Rightarrow I=4 \int \frac{t^{3}}{t^{2}+t} d t \\$
$\Rightarrow I=\int \frac{t^{3}+t^{2}-t^{2}}{t^{2}+t} d t\qquad\left[\text { we can write } t^{3}=t^{3}+t^{2}-t^{2}\right]$
$\begin{aligned} &\Rightarrow I=4 \int \frac{t\left(t^{2}+t\right)-t^{2}}{\left(t^{2}+t\right)} d t \\ & \end{aligned}$
$\Rightarrow I=4 \int\left(t-\frac{t^{2}}{t^{2}+t}\right) d t \\$
$\Rightarrow I=4 \int\left\{t-\frac{t^{2}+t}{t^{2}+t}-\frac{t}{t^{2}+t}\right\} d t$
$\begin{aligned} &\Rightarrow I=4 \int\left\{t-1-\frac{t}{t(t+1)}\right\} d t=4 \int\left\{t-1-\frac{1}{(t+1)}\right\} d t \\ & \end{aligned}$
$\Rightarrow I=4 \int t d t-4 \int t^{0} d t-4 \int \frac{1}{t+1} d t$
$\begin{aligned} &\Rightarrow I=4 \frac{t^{1+1}}{1+1}-4 \log |t+1|-4 t+c \\ & \end{aligned}$
$\Rightarrow I=4 \frac{t^{2}}{2}-4 t-4 \log |t+1|+c \\$
$\therefore I=2 \sqrt{x}-4 x^{\frac{1}{4}}+4 \log \left|1+x^{\frac{1}{4}}\right|+c$

Indefinite Integrals Excercise 18.10 Question 10

Answer: $3 x^{\frac{1}{3}}+3 \log \left|x^{\frac{1}{3}}-1\right|+c$
Hint: Use substitution method to solve this type of integral
Given: $\int \frac{1}{x^{\frac{1}{3}}\left(x^{\frac{1}{3}}-1\right)} d x$
Solution: let $I=\int \frac{1}{x^{\frac{1}{3}}\left(x^{\frac{1}{3}}-1\right)} d x$
Substitute $x=t^{3} \Rightarrow d x=3 t^{2} d t$ then
$\begin{aligned} I &=I=\int \frac{1}{t(t-1)} 3 t^{2} d t \quad\left(\because x^{\frac{1}{3}}=t\right) \\ & \end{aligned}$
$=3 \int \frac{t^{2}}{t(t-1)} d t=3 \int \frac{t}{t-1} d t \\$
$=3 \int \frac{t+1-1}{t-1} d t=3 \int\left(\frac{t-1}{t-1}+\frac{1}{t-1}\right)dt$
$\begin{aligned} &=3 \frac{t^{0+1}}{0+1}+3 \log |t-1|+c \qquad\left[\int \frac{1}{t} d t=\log |t|+c \& \int t^{n} d t=\frac{t^{n+1}}{n+1}+c\right] \\ & \end{aligned}$
$=3 t+3 \log |t-1|+c \\$
$\therefore I=3 x^{\frac{1}{3}}+3 \log \left|x^{\frac{1}{3}}-1\right|+c$

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