How can RD Sharma class 12 solutions help with Maths?

The key feature of these solutions is that each question is followed by a step-by-step explanation. There are no missed steps or shortcuts. As a result, if the student pays close attention, the answer will become self-explanatory. It also aids students in solidifying their concepts and acing their exams.

Can we revise using RD Sharma class 12 solutions?

Yes, having all of your answers in one place is useful for last-minute revision so you don't have to go looking for solutions. RD Sharma class 11 solutions for Mathematics will collect all of the answers to the exercises in one location for easy reference.

Are the solutions of Rd Sharma Class 12th Exercise 18.21 free?

Yes, RD Sharma Chapter 18 Exercise 18.21 Solutions are free of cost at Career360.

Is RD Sharma's book appropriate for exam preparation?

Yes, it can be a valuable resource for exam preparation and can help you achieve a high score. Solving more and more textbook solutions gives you enough practice.

How do you find the indefinite integral's value?

The process of finding the indefinite integral value of a function is also called integration or integrating f(x). This can be presented as: ∫f(x)dx = F(x) + C, where C is any real number. We generally use suitable formulas that help in getting the antiderivative of the given function. The result of the indefinite integral is a function.

RD Sharma Class 12 Exercise 18.21 Indefinite Integrals Solutions Maths - Download PDF Free Online

Schools
RD Sharma Solutions
RD Sharma Class 12 Exercise 18.21 Indefinite Integrals Solutions Maths - Download PDF Free Online

RD Sharma Class 12 Exercise 18.21 Indefinite Integrals Solutions Maths - Download PDF Free Online

Updated on 24 Jan 2022, 12:25 PM IST

Rd Sharma Class 12th Exercise 18.21 is very important because the RD Sharma Book has always been the best source for preparation. Every student studies from RD Sharma books and gets success. Mathematics being a complicated subject, needs best solutions for it which explain every bit of the concept to help students grasp it fast. Rd Sharma Solutions Rd Sharma Class 12th Exercise 18.21 has solved every problem of a student regarding differentiation and integration. This exercise has 18 questions based on quadratic equations, integrating the function, and some important formulae used for evaluating the integrals.

RD Sharma Class 12 Solutions Chapter18 Indefinite Integrals - Other Exercise

Indefinite Integrals Excercise:18.21

Indefinite Integrals Exercise 18.21 Question 1

Answer:$\sqrt{x^{2}+6 x+10}-3\left[\log \left|(x+3)+\sqrt{x^{2}+6 x+10}\right|\right]+c$
Given: $\int \frac{x}{\sqrt{x^{2}+6 x+10}} d x$
Hint: Simplify it and integrate it with the help of formula.
Solution: Let $I=\int \frac{x}{\sqrt{x^{2}+6 x+10}} d x$
$=\int \frac{x+3-3}{\sqrt{x^{2}+6 x+10}} d x$ ......................Adding +3 and -3 in numerator
$I=\frac{1}{2} \int \frac{2(x+3)}{\sqrt{x^{2}+6 x+10}} d x-3 \int \frac{d x}{\sqrt{x^{2}+6 x+10}}$
$I=I_{1}-I_{2}$ ..................(1)
Where
$\begin{aligned} &I_{1}=\frac{1}{2} \int \frac{2(x+3)}{\sqrt{x^{2}+6 x+10}} d x \\ &I_{2}=3 \int \frac{d x}{\sqrt{x^{2}+6 x+10}} \end{aligned}$
First solve $I_{1}$
Let
$\begin{aligned} &x^{2}+6 x+10=y \\ &(2 x+6) d x=d y \end{aligned}$ .........(2)
$I_{1}=\frac{1}{2} \int \frac{d y}{\sqrt{y}}=\frac{1}{2} \int(y)^{-\frac{1}{2}} d y$
$I_{1}=\frac{1}{2}\left[\frac{y^{\frac{1}{2}}}{\frac{1}{2}}\right]+c$
$I_{1}=\sqrt{y+c}$ From equation (2)
$I_{1}=\sqrt{x^{2}+6 x+10}+c$
Now $I_{2}=3 \int \frac{d x}{\sqrt{x^{2}+6 x+10}}$
$x^{2}+6 x+10=x^{2}+6 x+9-9+10$
$=\left ( x+3 \right )^{2}+1$
$I_{2}=3 \int \frac{d x}{\sqrt{(x+3)^{2}+1}}$
$I_{2}=3 \log \left|x+3+\sqrt{x^{2}+6 x+10}\right|+c$$\left[\int \frac{d x}{\sqrt{x^{2}+a^{2}}}=\log \left|x+\sqrt{x^{2}+a^{2}}\right|\right]$

Now, putting value of $I_{1}$ & $I_{2}$ in equation 1

$I=\sqrt{x^{2}+6 x+10}-3\left[\log \left|(x+3)+\sqrt{x^{2}+6 x+10}\right|\right]+c$

Indefinite Integrals Exercise 18.21 Question 2

Answer:$2 \sqrt{x^{2}+2 x-1}-\log \left|x+1+\sqrt{x^{2}+2 x-1}\right|+c$
Given:$\int \frac{2 x+1}{\sqrt{x^{2}+2 x-1}} d x$
Hint: Simplify it and solve it
Solution: Let $I=\int \frac{2 x+1}{\sqrt{x^{2}+2 x-1}} d x$
$I=\int \frac{2 x+2-1}{\sqrt{x^{2}+2 x-1}} d x$
$\begin{aligned} &I=\int \frac{2 x+2}{\sqrt{x^{2}+2 x-1}} d x-\int \frac{1}{\sqrt{x^{2}+2 x-1}} d x \\ &I=\int \frac{2 x+2}{\sqrt{x^{2}+2 x-1}} d x-\int \frac{1}{\sqrt{x^{2}+2 x+1-1-1}} d x \\ &I=\int \frac{2 x+2}{\sqrt{x^{2}+2 x-1}} d x-\int \frac{1}{\sqrt{(x+1)^{2}-2}} d x \end{aligned}$
Let
$x^{2}+2x-1=y$
$dx\left ( 2x+2 \right )=dy$
$I=\int \frac{d y}{\sqrt{y}}-\int \frac{1}{\sqrt{(x+1)^{2}-(\sqrt{2})^{2}}} d x$
$I=\frac{\sqrt{y}}{\frac{1}{2}}-\log \left|x+1+\sqrt{x^{2}+2 x-1}\right|+c$
$\begin{aligned} &{\left[\int \frac{d x}{\sqrt{x^{2}-a^{2}}}=\log \left|x+\sqrt{x^{2}-a^{2}}\right|+c\right]} \\ &I=2 \sqrt{x^{2}+2 x-1}-\log \left|x+1+\sqrt{x^{2}+2 x-1}\right|+c \end{aligned}$

Indefinite Integrals Exercise 18.21 Question 3

Answer:$-\sqrt{4+5 x-x^{2}}+\frac{7}{2} \sin ^{-1}\left(\frac{2 x-5}{\sqrt{41}}\right)+c$
Given: $\int \frac{x+1}{\sqrt{-x^{2}+5 x+4}} d x$
Hint: Simplify it and solve it
Solution: $I=\int \frac{x+1}{\sqrt{-x^{2}+5 x+4}} d x$
$I=\int \frac{x+1}{\sqrt{4-\left(x^{2}-5 x\right)}} d x=\int \frac{x+1}{\sqrt{4+\left(\frac{5}{2}\right)^{2}-\left(x^{2}-5 x+\left(\left(\frac{5}{2}\right)^{2}\right)\right.}} d x$
$I=\int \frac{x+1}{\sqrt{\frac{41}{4}-\left(x-\frac{5}{2}\right)^{2}}} d x$
$I=\frac{1}{2} \int \frac{2 x-5+7}{\sqrt{\frac{41}{4}-\left(x-\frac{5}{2}\right)^{2}}} d x$
$=\frac{1}{2} \int \frac{2 x-5}{\sqrt{\frac{41}{4}-\left(x-\frac{5}{2}\right)^{2}}} d x+\frac{7}{2} \int \frac{d x}{\sqrt{\frac{41}{4}-\left(x-\frac{5}{2}\right)^{2}}}$
$\frac{41}{4}-\left(x-\frac{5}{2}\right)^{2}=y$
$\begin{aligned} &-2\left(x-\frac{5}{2}\right) d x=d y \\ &(2 x-5) d x=-d y \end{aligned}$
$\Rightarrow I=\frac{1}{2} \int \frac{-d y}{\sqrt{y}}+\frac{7}{2} \int \frac{d x}{\sqrt{\frac{41}{4}-\left(x-\frac{5}{2}\right)^{2}}}$
$\Rightarrow I=-\frac{1}{2}\left[\frac{(y)^{\frac{1}{2}}}{\frac{1}{2}}\right]+\frac{7}{2}\left(\sin ^{-1}\left(\frac{x-\frac{5}{2}}{\frac{\sqrt{41}}{2}}\right)\right)+c \quad\left[\int \frac{1}{\sqrt{a^{2}-x^{2}}}=\sin ^{-1}\left(\frac{x}{a}\right)+c\right]$
$\begin{aligned} &\Rightarrow I=-\left[\frac{41}{4}-\left(x-\frac{5}{2}\right)^{2}\right]^{\frac{1}{2}}+\frac{7}{2} \sin ^{-1}\left(\frac{2 x-5}{\sqrt{41}}\right)+c \\ &\Rightarrow I=-\sqrt{5 x+4-x^{2}}+\frac{7}{2} \sin ^{-1}\left(\frac{2 x-5}{\sqrt{41}}\right)+c \end{aligned}$

Indefinite Integrals Exercise 18.21 Question 4

Answer: $2 \sqrt{3 x^{2}-5 x+1}+c$
Given:$\int \frac{6 x-5}{\sqrt{3 x^{2}-5 x+1}} d x$
Hint: Simplify it and solve it
Solution:
Let $I=\int \frac{6 x-5}{\sqrt{3 x^{2}-5 x+1}} d x$
Let $3 x^{2}-5 x+1=y$
$\Rightarrow(6 x-5) d x=d y$
$I=\int \frac{d y}{\sqrt{y}}$
$I=\frac{\sqrt{y}}{\frac{1}{2}}+c$
$\begin{aligned} &I=2 \sqrt{y}+c \\ &I=2 \sqrt{3 x^{2}-5 x+1}+c \end{aligned}$
(From equation (1))

Indefinite Integrals Exercise 18.21 Question 5

Answer: $-3 \sqrt{5-x^{2}-2 x}-2 \sin ^{-1}\left(\frac{x+1}{\sqrt{6}}\right)+c$
Given: $\int \frac{3 x+1}{\sqrt{5-x^{2}-2 x}} d x$
Hint: Simplify it and then integrate
Solution:
$\begin{aligned} &I=\int \frac{3 x+1}{\sqrt{5-x^{2}-2 x}} d x \\ &I=\int \frac{3 x+1}{\sqrt{6-\left(x^{2}+2 x+1\right)}} d x \\ &I=\int \frac{3 x+1}{\sqrt{6-(x+1)^{2}}} d x \\ &I=3 \int \frac{x+\frac{1}{3}}{\sqrt{6-(x+1)^{2}}} \end{aligned}$
$=-\frac{3}{2} \int \frac{-2 x-\frac{2}{3}}{\sqrt{6-(x+1)^{2}}}$ $\text { (multiplying and dividing numerator by-2 }$
$\begin{aligned} &I=-\frac{3}{2} \int \frac{-2 x-2-\frac{2}{3}+2}{\sqrt{6-(x+1)^{2}}} d x \\ &I=-\frac{3}{2} \int \frac{-2 x-2}{\sqrt{5-x^{2}-2 x}}+-\frac{3}{2}\left(\frac{4}{3}\right) \int \frac{1}{\sqrt{6-(x+1)^{2}}} d x \end{aligned}$
$I=I_{1}+I_{2}$ .................(1)
$I_{1}=-\frac{3}{2} \int \frac{-2 x-2}{\sqrt{5-x^{2}-2 x}} d x ; I_{2}=-2 \int \frac{1}{\sqrt{6-(x+1)^{2}}} d x$
Now,
Let
$5-x^{2}-2 x=y$ ..........(2)
$(-2 x-2) d x=d y$
$\begin{aligned} &\Rightarrow I_{1}=-\frac{3}{2} \int \frac{d x}{\sqrt{y}} \\ &I_{1}=-\frac{3}{2}\left(\frac{\sqrt{y}}{\frac{1}{2}}\right)+c \end{aligned}$ From Equation (2)
$I_{1}=-3 \sqrt{5-x^{2}-2 x}+c$
Now,$I_{2}=-2 \int \frac{1}{\sqrt{6-(x+1)^{2}}} d x$
$I_{2}=-2 \sin ^{-1}\left(\frac{x+1}{\sqrt{6}}\right)+c$
$\left[\int \frac{1}{\sqrt{a^{2}-x^{2}}} d x=\sin ^{-1} \frac{x}{a}+c\right]$
Putting value of $I_{1}$&$I_{2}$ in equation (1) and we get
$I=-3 \sqrt{5-x^{2}-2 x}-2 \sin ^{-1}\left(\frac{x+1}{\sqrt{6}}\right)+c$

Indefinite Integrals Exercise 18.21 Question 6

Answer: $-\sqrt{8+x-x^{2}}+\frac{1}{2} \sin ^{-1}\left(\frac{2 x-1}{\sqrt{33}}\right)+c$
Given: $\int \frac{x}{\sqrt{8+x-x^{2}}} d x$
Hint: Simplify the given function
Solution:
$\begin{aligned} &I=\int \frac{x}{\sqrt{8+x-x^{2}}} d x \\ &I=-\frac{1}{2} \int \frac{-2 x+1-1}{\sqrt{8+x-x^{2}}} d x \\ &I=-\frac{1}{2} \int \frac{-2 x+1}{\sqrt{8+x-x^{2}}} d x+\frac{1}{2} \int \frac{1}{\sqrt{8+x-x^{2}}} d x \end{aligned}$
$I=I_{1}+I_{2}$ ...................(1)
$I_{1}=-\frac{1}{2} \int \frac{-2 x+1}{\sqrt{8+x-x^{2}}} d x \text { and } I_{2}=\frac{1}{2} \int \frac{1}{\sqrt{8+x-x^{2}}} d x$
$I_{1}=-\frac{1}{2} \int \frac{-2 x+1}{\sqrt{8+x-x^{2}}} d x$
Let
$8+x-x^{2}=y$ .............(2)
$(1-2 x) d x=d y$
$I_{1}=-\frac{1}{2} \int \frac{d y}{\sqrt{y}}=-\frac{1}{2}\left(\frac{\sqrt{y}}{\frac{1}{2}}\right)+c$
$\begin{aligned} &I_{1}=-\sqrt{y}+c \\ &I_{1}=-\sqrt{8+x-x^{2}}+c \end{aligned}$ (From Equation 2)
Now,$I_{2}=\frac{1}{2} \int \frac{1}{\sqrt{8+x-x^{2}}} d x$
$I_{2}=\frac{1}{2} \int \frac{1}{\sqrt{8-\left(x^{2}-x+\frac{1}{4}\right)+\frac{1}{4}}} d x$
$I_{2}=\frac{1}{2} \int \frac{d x}{\sqrt{\frac{33}{4}-\left(x-\frac{1}{2}\right)^{2}}}$
$I_{2}=\frac{1}{2} \sin ^{-1}\left(\frac{2 x-1}{\sqrt{33}}\right)+c$
Putting $I_{1}$ & $I_{2}$ in equation (1)
$I=-\sqrt{8+x-x^{2}}+\frac{1}{2} \sin ^{-1}\left(\frac{2 x-1}{\sqrt{33}}\right)+c$

Indefinite Integrals Exercise 18.21 Question 7

Answer: $\sqrt{x^{2}+2 x-1}+\log \left|x+1+\sqrt{x^{2}+2 x-1}\right|+c$
Given: $\int \frac{x+2}{\sqrt{x^{2}+2 x-1}} d x$
Hint: Simplify the given function
Solution:
$\begin{aligned} &I=\int \frac{x+2}{\sqrt{x^{2}+2 x-1}} d x \\ &I=\frac{1}{2} \int \frac{2 x+4}{\sqrt{x^{2}+2 x-1}} d x \\ &I=\frac{1}{2} \int \frac{2 x+2}{\sqrt{x^{2}+2 x-1}} d x+\frac{2}{2} \int \frac{1}{\sqrt{x^{2}+2 x-1}} d x \\ &I=\frac{1}{2} \int \frac{2 x+2}{\sqrt{x^{2}+2 x-1}} d x+1 \int \frac{1}{\sqrt{x^{2}+2 x+1-2}} d x \\ &I=\frac{1}{2} \int \frac{2 x+2}{\sqrt{x^{2}+2 x-1}} d x+\int \frac{1}{\sqrt{(x+1)^{2}-(\sqrt{2})^{2}}} d x \end{aligned}$
$I=\frac{1}{2}\left[\frac{\sqrt{x^{2}+2 x-1}}{\frac{1}{2}}\right]+\log \left|x+1+\sqrt{x^{2}+2 x-1}\right|+c$
$\left[\begin{array}{l} U \sin g \\ (f(x))^{n} f^{1}(x) d x=\frac{[f(x)]^{n+1}}{n+1} \\ \int \frac{1}{\sqrt{x^{2}-a^{2}}} d x=\log \left|x+\sqrt{x^{2}-a^{2}}\right|+c \end{array}\right]$
$I=\sqrt{x^{2}+2 x-1}+\log \left|x+1+\sqrt{x^{2}+2 x-1}\right|+c$

Indefinite Integrals Exercise 18.21 Question 8

Answer: $\sqrt{x^{2}-1}+2 \log \left|x+\sqrt{x^{2}-1}\right|+c$
Given: $\int \frac{x+2}{\sqrt{x^{2}-1}} d x$
Hint: Simplify the given function
Solution: $I=\int \frac{x+2}{\sqrt{x^{2}-1}} d x$
$\begin{aligned} &I=\frac{1}{2} \int \frac{2 x}{\sqrt{x^{2}-1}} d x+2 \int \frac{1}{\sqrt{x^{2}-1}} d x \\ &I=\frac{1}{2}\left[\frac{x^{2}-1}{\frac{1}{2}}\right]+2 \log \left|x+\sqrt{x^{2}-1}\right|+c \end{aligned}$
$\left[\begin{array}{l} U \sin g \\ \int(f(x))^{n} f^{1}(x) d x=\frac{f(x)^{n+1}}{n+1}+c \\ \int \frac{1}{\sqrt{x^{2}-a^{2}}} d x=\log \left|x+\sqrt{x^{2}-a^{2}}\right|+c \end{array}\right]$
$I=\sqrt{x^{2}-1}+2 \log \left|x+\sqrt{x^{2}-1}\right|+c$

Indefinite Integrals Exercise 18.21 Question 9

Answer: $\sqrt{x^{2}+1}-\log \left|x+\sqrt{x^{2}+1}\right|+c$
Given: $\int \frac{x-1}{\sqrt{x^{2}+1}} d x$
Hint: Simplify the given function
Solution:
$\begin{aligned} &I=\int \frac{x-1}{\sqrt{x^{2}+1}} d x \\ &I=\frac{1}{2} \int \frac{2 x}{\sqrt{x^{2}+1}} d x-\int \frac{1}{\sqrt{x^{2}+1}} d x \\ &I=\frac{1}{2}\left[\frac{\sqrt{x^{2}+1}}{\frac{1}{2}}\right]-\log \left|x+\sqrt{x^{2}+1}\right|+c \end{aligned}$
$\left[\begin{array}{l} U \sin g \\ \int(f(x))^{n} f^{1}(x) d x=\frac{[f(x)]^{n+1}}{n+1}+c \\ \int \frac{1}{\sqrt{x^{2}+a^{2}}} d x=\log \left|x+\sqrt{x^{2}+a^{2}}\right|+c \end{array}\right]$
$I=\sqrt{x^{2}+1}-\log \left|x+\sqrt{x^{2}+1}\right|+c$

Indefinite Integrals Exercise 18.21 Question 10

Answer: $\sqrt{x^{2}+x+1}-\frac{1}{2} \log \left|\frac{2 x+1}{2}+\sqrt{x^{2}+x+1}\right|+c$
Given: $\int \frac{x}{\sqrt{x^{2}+x+1}} d x$
Hint: Simplify the given $(f(x))^{n}$
Solution:
$\begin{aligned} &I=\int \frac{x}{\sqrt{x^{2}+x+1}} d x \\ &I=\frac{1}{2} \int \frac{2 x}{\sqrt{x^{2}+x+1}} d x \\ &I=\frac{1}{2} \int \frac{2 x+1-1}{\sqrt{x^{2}+x+1}} d x \\ &I=\frac{1}{2} \int \frac{2 x+1}{\sqrt{x^{2}+x+1}} d x-\frac{1}{2} \int \frac{1}{\sqrt{x^{2}+x+1}} d x \\ &I=\frac{1}{2} \int \frac{2 x+1}{\sqrt{x^{2}+x+1}} d x-\frac{1}{2} \int \frac{1}{\sqrt{\left(x^{2}+x+\frac{1}{4}\right)+1-\frac{1}{4}}} d x \end{aligned}$
$\begin{aligned} &I=\frac{1}{2} \int \frac{2 x+1}{\sqrt{x^{2}+x+1}} d x-\frac{1}{2} \int \frac{1}{\sqrt{\left(x+\frac{1}{2}\right)^{2}+\frac{3}{4}}} d x \\ &I=\frac{1}{2}\left[\frac{\sqrt{x^{2}+x+1}}{\frac{1}{2}}\right]-\frac{1}{2} \log \left|x+\frac{1}{2}+\sqrt{x^{2}+x+1}\right|+c \end{aligned}$
$\left[\begin{array}{l} U \sin g \\ \int(f(x))^{n} f^{1}(x) d x=\frac{[f(x)]^{n+1}}{n+1}+c \\ \int \frac{1}{\sqrt{x^{2}+a^{2}}} d x=\log \left|x+\sqrt{x^{2}+a^{2}}\right|+c \end{array}\right]$
$I=\sqrt{x^{2}+x+1}-\frac{1}{2} \log \left|\frac{2 x+1}{2}+\sqrt{x^{2}+x+1}\right|+c$

Indefinite Integrals Exercise 18.21 Question 11

Answer: $\sqrt{x^{2}+1}+\log \left|x+\sqrt{x^{2}+1}\right|+c$
Given: $\int \frac{x+1}{\sqrt{x^{2}+1}} d x$
Hint: Simplify the given integration and solve it
Solution:
$\begin{aligned} &I=\int \frac{x+1}{\sqrt{x^{2}+1}} d x \\ &I=\int \frac{x}{\sqrt{x^{2}+1}} d x+\int \frac{1}{\sqrt{x^{2}+1}} d x \end{aligned}$
$\begin{aligned} &I=\frac{1}{2} \int \frac{2 x}{\sqrt{x^{2}+1}} d x+\log \left|x+\sqrt{x^{2}+1}\right|+c \\ &I=I_{1}+\log \left|x+\sqrt{x^{2}+1}\right|+c \\ &I_{1}=\frac{1}{2} \int \frac{2 x}{\sqrt{x^{2}+1}} d x \end{aligned}$
$\begin{aligned} &\text { Let } \\ &x^{2}+1=y \\ &2 x d x=d y \end{aligned}$
$\begin{aligned} &I_{1}=\frac{1}{2} \int \frac{d y}{\sqrt{y}}=\frac{1}{2}\left(\frac{\sqrt{y}}{\frac{1}{2}}\right)+c \\ &I_{1}=\sqrt{y}+c \\ &I_{1}=\sqrt{x^{2}+1}+c \\ &I=\sqrt{x^{2}+1}+\log \left|x+\sqrt{x^{2}+1}\right|+c \end{aligned}$

Indefinite Integrals Exercise 18.21 Question 12

Answer: $2 \sqrt{x^{2}+2 x+5}+3 \log \left|x+1+\sqrt{x^{2}+2 x+5}\right|+c$
Given: $\int \frac{2 x+5}{\sqrt{x^{2}+2 x+5}} d x$

Hint: Simplify the given (f(x))ⁿ
Solution:
$\begin{aligned} &I=\int \frac{2 x+5}{\sqrt{x^{2}+2 x+5}} d x \\ &I=\int \frac{2 x+2+3}{\sqrt{x^{2}+2 x+5}} d x \\ &I=\int \frac{2 x+2}{\sqrt{x^{2}+2 x+5}} d x+3 \int \frac{1}{\sqrt{x^{2}+2 x+5}} d x \end{aligned}$
$I=I_{1}+I_{2}$ .......................(1)
Where
$I_{1}=\int \frac{2 x+2}{\sqrt{x^{2}+2 x+5}} d x \& I_{2}=3 \int \frac{1}{\sqrt{x^{2}+2 x+5}} d x$
Now,Let
$\begin{aligned} &x^{2}+2 x+5=y \\ &(2 x+2) d x=d y \end{aligned}$ ..........................(2)
$\begin{aligned} &I_{1}=2 \sqrt{y}+c \\ &I_{1}=2 \sqrt{x^{2}+2 x+5}+c \end{aligned}$ ( From equation 2)
Now, $I_{2}=3 \int \frac{1}{\sqrt{x^{2}+2 x+4+1}} d x$
$I_{2}=3 \int \frac{1}{\sqrt{(x+1)^{2}+2^{2}}} d x$ $\left[\int \frac{1}{\sqrt{x^{2}+a^{2}}} d x=\log \left|x+\sqrt{x^{2}+a^{2}}\right|+c\right]$
$I_{2}=3 \log \left|x+1+\sqrt{x^{2}+2 x+5}\right|+c$
Putting value of $I_{1}$ & $I_{2}$in equation (1)
$I=2 \sqrt{x^{2}+2 x+5}+3 \log \left|x+1+\sqrt{x^{2}+2 x+5}\right|+c$

Indefinite Integrals Exercise 18.21 Question 13

Answer: $-3 \sqrt{5-x^{2}-2 x}-2 \sin ^{-1}\left(\frac{x+1}{\sqrt{6}}\right)+c$
Given: $\int \frac{3 x+1}{\sqrt{5-x^{2}-2 x}} d x$
Hint: Simplify the given Function
Solution:
$\begin{aligned} &I=\int \frac{3 x+1}{\sqrt{5-x^{2}-2 x}} d x \\ &I=3 \int \frac{x+\frac{1}{3}}{\sqrt{5-x^{2}-2 x}} d x \end{aligned}$
Multiplying and dividing the numerator by -2,
$\begin{aligned} I &=\frac{-3}{2} \int \frac{-2 x-\frac{2}{3}}{\sqrt{5-x^{2}-2 x}} d x \\ I &=\frac{-3}{2} \int \frac{-2 x-2+\frac{4}{3}}{\sqrt{5-x^{2}-2 x}} d x \\ I &=\frac{-3}{2} \int \frac{-2 x-2}{\sqrt{5-x^{2}-2 x}} d x-2 \int \frac{1}{\sqrt{5-x^{2}-2 x}} d x \\ I &=I_{1}+I_{2} \\ I_{1} &=\frac{-3}{2} \int \frac{-2 x-2}{\sqrt{5-x^{2}-2 x}} d x ; I_{2}=-2 \int \frac{1}{\sqrt{5-x^{2}-2 x}} d x \end{aligned}$
Now, $I_{1}=\frac{-3}{2} \int \frac{-2 x-2}{\sqrt{5-x^{2}-2 x}} d x$
Let
$\begin{aligned} &5-x^{2}-2 x=y \\ &(-2-2 x) d x=d y \end{aligned}$
$\Rightarrow I_{1}=\frac{-3}{2} \int \frac{d y}{\sqrt{y}}=\frac{-3}{2}\left(\frac{\sqrt{y}}{\frac{1}{2}}\right)+c$
$\begin{aligned} &I_{1}=-3 \sqrt{y}+c \\ &I_{1}=-3 \sqrt{5-x^{2}-2 x}+c \end{aligned}$
Now, $I_{2}=-2 \int \frac{1}{\sqrt{5-x^{2}-2 x}} d x$
$\begin{aligned} &I_{2}=-2 \int \frac{1}{\sqrt{5-\left(x^{2}+2 x+1\right)+1}} d x \\ &I_{2}=-2 \int \frac{1}{\sqrt{6-(x+1)^{2}}} d x \\ &I_{2}=-2 \sin ^{-1}\left[\frac{x+1}{\sqrt{6}}\right]+c \end{aligned}$
Putting $I_{1}$ & $I_{2}$ in equation (1) we get
$I=-3 \sqrt{5-x^{2}-2 x}-2 \sin ^{-1}\left(\frac{x+1}{\sqrt{6}}\right)+c$

Indefinite Integrals Exercise 18.21 Question 14

Answer: $\sin ^{-1} x+\sqrt{1-x^{2}}+c$
Given: $\int \sqrt{\frac{1-x}{1+x}} d x$
Hint: Simplify the given function
Solution:
$\begin{aligned} I &=\int \sqrt{\frac{1-x}{1+x}} d x \\ I &=\int \sqrt{\frac{1-x}{1+x} \times \frac{1-x}{1-x}} d x \text { (rationalizing ta e denominator) } \\ &=\int \sqrt{\frac{(1-x)^{2}}{1-x^{2}}} d x \quad\left[(a+b)(a-b)=\left(a^{2}-b^{2}\right)\right] \\ I &=\int \frac{1-x}{\sqrt{1-x^{2}}} d x \end{aligned}$
$\begin{aligned} &I=\int \frac{1}{\sqrt{1-x^{2}}} d x-\frac{1}{2} \int \frac{2 x}{\sqrt{1-x^{2}}} d x \\ &I=\sin ^{-1} x+\frac{1}{2}\left(\frac{\sqrt{1-x^{2}}}{\frac{1}{2}}\right)+c \end{aligned}$
$\begin{aligned} &{\left[\begin{array}{c} \int \frac{d x}{\sqrt{a^{2}-x^{2}}}=\sin ^{-1}\left(\frac{x}{a}\right)+c \\ \int(f(x))^{n} f^{1}(x) d x=\frac{[f(x)]^{n+1}}{n+1}+c \end{array}\right]} \\ &I=\sin ^{-1} x+\sqrt{1-x^{2}}+c \end{aligned}$

Indefinite Integrals Exercise 18.21 Question 15

Answer: $2 \sqrt{x^{2}+4 x+3}-3 \log \left|x+2+\sqrt{x^{2}+4 x+3}\right|+c$
Given: $\int \frac{2 x+1}{\sqrt{x^{2}+4 x+3}} d x$
Hint: Simplify the given function
Solution:
$\begin{aligned} &I=\int \frac{2 x+1}{\sqrt{x^{2}+4 x+3}} d x \\ &I=\int \frac{2 x+4-3}{\sqrt{x^{2}+4 x+3}} d x \\ &I=\int \frac{2 x+4}{\sqrt{x^{2}+4 x+3}} d x-\int \frac{3}{\sqrt{x^{2}+4 x+3}} d x \\ &I=\int \frac{2 x+4}{\sqrt{x^{2}+4 x+3}} d x-\int \frac{3}{\sqrt{x^{2}+4 x+4-4+3}} d x \\ &I=\int \frac{2 x+4}{\sqrt{x^{2}+4 x+3}} d x-\int \frac{3}{\sqrt{(x+2)^{2}-1}} d x \end{aligned}$
$\begin{aligned} &I=I_{1}-3 \log \left|x+2+\sqrt{x^{2}+4 x+3}\right|+c \quad \ldots\left[\int \frac{1}{\sqrt{x^{2}-a^{2}}} d x=\log \left|x+\sqrt{x^{2}-a^{2}}\right|+c\right] \\ &I_{1}=\int \frac{2 x+4}{\sqrt{x^{2}+4 x+3}} d x \end{aligned}$
Let,
$\begin{aligned} &x^{2}+4 x+3=y \\ &(2 x+4) d x=d y \\ &I_{1}=\int \frac{d y}{\sqrt{y}}=\frac{\sqrt{y}}{\frac{1}{2}}+c \\ &I_{1}=2 \sqrt{x^{2}+4 x+3}+c \\ &I=2 \sqrt{x^{2}+4 x+3}-3 \log \left|x+2+\sqrt{x^{2}+4 x+3}\right|+c \end{aligned}$

Indefinite Integrals Exercise 18.21 Question 16

Answer: $2 \sqrt{x^{2}+4 x+5}-\log \left|x+2+\sqrt{x^{2}+4 x+5}\right|+c$
Given: $\int \frac{2 x+3}{\sqrt{x^{2}+4 x+5}} d x$
Hint: Simplify the given function
Solution:
$\begin{aligned} &I=\int \frac{2 x+3}{\sqrt{x^{2}+4 x+5}} d x \\ &I=\int \frac{2 x+4-1}{\sqrt{x^{2}+4 x+5}} d x \\ &I=\int \frac{2 x+4}{\sqrt{x^{2}+4 x+5}} d x-\int \frac{1}{\sqrt{x^{2}+4 x+4-4+5}} d x \\ &I=\int \frac{2 x+4}{\sqrt{x^{2}+4 x+5}} d x-\int \frac{d x}{\sqrt{(x+2)^{2}+1}} \\ &I=\int \frac{2 x+4}{\sqrt{x^{2}+4 x+5}} d x-\log \left|x+2+\sqrt{x^{2}+4 x+5}\right|+c \end{aligned}$
Using....................$\left[\int \frac{1}{\sqrt{x^{2}+a^{2}}} d x=\log \left|x+\sqrt{x^{2}+a^{2}}\right|+c\right]$
Let
$\begin{aligned} &x^{2}+4 x+5=y \\ &(2 x+4) d x=d y \end{aligned}$
$\Rightarrow \int \frac{2 x+4}{\sqrt{x^{2}+4 x+5}} d x=\int \frac{d y}{\sqrt{y}}=\frac{\sqrt{y}}{\frac{1}{2}}+c$
$\begin{aligned} &=2 \sqrt{y}+c \\ &=2 \sqrt{x^{2}+4 x+5}+c \end{aligned}$
$I=2 \sqrt{x^{2}+4 x+5}-\log \left|x+2+\sqrt{x^{2}+4 x+5}\right|+c$

Indefinite Integrals Exercise 18.21 Question 17

Answer: $5 \sqrt{x^{2}+4 x+10}-7 \log \left|x+2+\sqrt{x^{2}+4 x+10}\right|$
Given: $\int \frac{5 x+3}{\sqrt{x^{2}+4 x+10}} d x$
Hint: Simplify the given function
Solution:
$\begin{aligned} &I=\int \frac{5 x+3}{\sqrt{x^{2}+4 x+10}} d x\\ &I=5 \int \frac{x+\frac{3}{5}}{\sqrt{x^{2}+4 x+10}} d x \text { multiplying and dividing by } 2 \text { in numerator, }\\ &I=\frac{5}{2} \int \frac{2 x+\frac{6}{5}}{\sqrt{x^{2}+4 x+10}} d x\\ &I=\frac{5}{2} \int \frac{2 x+4-\frac{14}{5}}{\sqrt{x^{2}+4 x+10}} d x \end{aligned}$
$\begin{aligned} &I=\frac{5}{2} \int \frac{2 x+4}{\sqrt{x^{2}+4 x+10}} d x-7 \int \frac{1}{\sqrt{x^{2}+4 x+10}} d x \\ &I=\frac{5}{2} \int \frac{2 x+4}{\sqrt{x^{2}+4 x+10}} d x-7 \int \frac{1}{\sqrt{x^{2}+4 x+4+6}} d x \\ &I=\frac{5}{2} \int \frac{2 x+4}{\sqrt{x^{2}+4 x+10}} d x-7 \int \frac{1}{\sqrt{(x+2)^{2}+(\sqrt{6})^{2}}} d x \\ &I=\frac{5}{2} \int \frac{2 x+4}{\sqrt{x^{2}+4 x+10}} d x-7 \log \left|x+2+\sqrt{x^{2}+4 x+10}\right| \end{aligned}$
using....................$\left[\int \frac{1}{\sqrt{x^{2}+a^{2}}} d x=\log \left|x+\sqrt{x^{2}+a^{2}}\right|+c\right]$
Let
$\begin{aligned} &x^{2}+4 x+10=y \\ &(2 x+4) d x=d y \end{aligned}$
$\Rightarrow \int \frac{2 x+4}{\sqrt{x^{2}+4 x+10}} d x=\int \frac{d y}{\sqrt{y}}=\frac{\sqrt{y}}{\frac{1}{2}}+c$
$\begin{aligned} &=2 \sqrt{y}+c \\ &=2 \sqrt{x^{2}+4 x+10}+c \end{aligned}$
$I=5 \sqrt{x^{2}+4 x+10}-7 \log \left|x+2+\sqrt{x^{2}+4 x+10}\right|$

Indefinite Integrals Exercise 18.21 Question 18

Answer: $\sqrt{x^{2}+2 x+3}+\log \left|x+1+\sqrt{x^{2}+2 x+3}\right|+c$
Given: $\int \frac{x+2}{\sqrt{x^{2}+2 x+3}} d x$
Hint: Simplify the given function
Solution:
$\begin{aligned} &I=\int \frac{x+2}{\sqrt{x^{2}+2 x+3}} d x \\ &I=\frac{1}{2} \int \frac{2 x+4}{\sqrt{x^{2}+2 x+3}} d x \\ &I=\frac{1}{2} \int \frac{2 x+2}{\sqrt{x^{2}+2 x+3}} d x+\frac{1}{2} \int \frac{2}{\sqrt{x^{2}+2 x+3}} d x \\ &I=\frac{1}{2} \int \frac{2 x+2}{\sqrt{x^{2}+2 x+3}} d x+\int \frac{1}{\sqrt{x^{2}+2 x+1+2}} d x \\ &I=\frac{1}{2} \int \frac{2 x+2}{\sqrt{x^{2}+2 x+3}} d x+\int \frac{1}{\sqrt{(x+1)^{2}+2}} d x \end{aligned}$
$I=\frac{1}{2} \int \frac{2 x+2}{\sqrt{x^{2}+2 x+3}} d x+\log \left|(x+1)+\sqrt{x^{2}+2 x+3}\right|+c\left|\int \frac{1}{\sqrt{x^{2}+a^{2}}} d x=\log \right| x+\sqrt{x^{2}+a^{2}}|+c|$
Let
$\begin{aligned} &x^{2}+2 x+3=y \\ &(2 x+2) d x=d y \end{aligned}$
$\Rightarrow \int \frac{2 x+2}{\sqrt{x^{2}+2 x+3}} d x=\int \frac{d y}{\sqrt{y}}=\frac{\sqrt{y}}{\frac{1}{2}}+c$
$\begin{aligned} &=2 \sqrt{y}+c \\ &=2 \sqrt{x^{2}+2 x+3}+c \end{aligned}$
$\begin{aligned} &\Rightarrow I=\frac{1}{2}\left(2 \sqrt{x^{2}+2 x+3}\right)+\log \left|x+1+\sqrt{x^{2}+2 x+3}\right|+c \\ &\Rightarrow I=\sqrt{x^{2}+2 x+3}+\log \left|x+1+\sqrt{x^{2}+2 x+3}\right|+c \end{aligned}$

Rd Sharma Class 12th Exercise 18.21 solutions will help every student because of the plenty of benefits. Some benefits of these solutions are listed below:

Experts have created it with full focus:

These solutions are created by a team of experts who have focused on concept building and grades of students. The team discusses every question in detail, gives solutions and provides numerous ways for one question. Therefore, a student will always be able to learn the concept behind every question in mathematics after going for Rd Sharma Class 12 Chapter 18 Exercise 18.21 solutions.

Best material to prepare for board and competitive exams:

JEE Main Highest Scoring Chapters & Topics

Focus on high-weightage topics with this eBook and prepare smarter. Gain accuracy, speed, and a better chance at scoring higher.

Download E-book

These solutions help students to understand the concept behind every question and will be able to solve it. The students will be ready for any exam with RD Sharma Class 12 Solutions Indefinite Integrals ex 18.21 .

Questions from NCERT textbook

The solutions help a student do the homework from teachers as they usually follow NCERT for questions. The student will be able to do it effectively as these solutions are mostly based on NCERT questions. These solutions help students achieve good grades in mid term exams also.

Numerous approaches to solve a single question:

RD Sharma Class 12 Solutions Chapter 10 ex 18.21 helps students find alternative ways to solve a single problem in mathematics due to the great minds behind its creation. A student finds it easy to choose and understand the requirements of the problem.

Great performance is guaranteed

The solutions help a student perform exceptionally in exams. RD Sharma Class 12 Solutions Chapter 18 ex 18.21 covers all the important questions that come in exams and give various ways to solve them. A student will be able to write exams with confidence and a victory smile.

Free of cost

Students will find all these solutions free of cost at Career360. Career360 is the best website to visit and learn the concepts due to the reasons stated above. Thousands of students have gained a lot of knowledge and excellent marks from these solutions making Career360 number one choice for every student.

RD Sharma Chapter-wise Solutions