RD Sharma Class 12 Exercise 18.15 Indefinite Integrals Solutions Maths - Download PDF Free Online

# RD Sharma Class 12 Exercise 18.15 Indefinite Integrals Solutions Maths - Download PDF Free Online

Edited By Kuldeep Maurya | Updated on Jan 24, 2022 12:15 PM IST

It is a well-known fact that RD Sharma class 12th exercise 18.15 is a highly recommended NCERT solution. Students and teachers from across the Indian subcontinent have expressed their trust in RD Sharma Solutions, which has become a popular name in the education industry. The RD Sharma class 12 chapter 18 exercise 18.15 is especially favoured by students as it contains answers to a challenging section from the maths book.

The RD Sharma class 12 solutions Indefinite Integrals 18.15 will cover the chapter Indefinite Integrals which is complex and pretty lengthy. The chapter will test stunning on the concepts of Integrals, Graphs of indefinite integrals, Indefinite integrals of common functions, solving the quadratic equation etc. The exercise 18.15 part has a total of 6 questions. The RD Sharma class 12th exercise 18.15 will provide students with easy answers to all 6 of them.

## Indefinite Integrals Excercise: 18.15

Indefinite Integrals Exercise 18.15 Question 1

$\frac{1}{8} \log \left|\frac{2 x+1}{2 x+5}\right|+C$
Hint:
To solve this problem use special integration formula
Given:
$\int \frac{1}{4 x^{2}+12 x+5} d x$
Solution:
Let $I=\int \frac{1}{4 x^{2}+12 x+5} d x=\frac{1}{4} \int \frac{1}{x^{2}+\frac{12 x}{4}+\frac{5}{4}} d x$
\begin{aligned} &=\frac{1}{4} \int \frac{1}{x^{2}+3 x+\frac{5}{4}} d x=\frac{1}{4} \int \frac{d x}{x^{2}+2 \cdot x \cdot \frac{3}{2}+\left(\frac{3}{2}\right)^{2}-\left(\frac{3}{2}\right)^{2}+\frac{5}{4}} \\\\ &=\frac{1}{4} \int \frac{1}{\left(x+\frac{3}{2}\right)^{2}-\frac{9}{4}+\frac{5}{4}} d x=\frac{1}{4} \int \frac{1}{\left(x+\frac{3}{2}\right)^{2}-\left(\frac{9-5}{4}\right)} d x \end{aligned}

Put $x+\frac{3}{2}=t \Rightarrow d x=d t$

Then $I=\frac{1}{4} \int \frac{1}{t^{2}-1^{2}} d t$
$=\frac{1}{4} \cdot \frac{1}{2 \times 1} \log \left|\frac{t-1}{t+1}\right|+C$ $\quad\left[\because \int \frac{1}{x^{2}-a^{2}} d x=\frac{1}{2 a} \log \left|\frac{x-a}{x+a}\right|+C\right]$
$=\frac{1}{4} \times \frac{1}{2} \log \left|\frac{x+\frac{3}{2}-1}{x+\frac{3}{2}+1}\right|+C \text { }$ $\left[\because t=x+\frac{3}{2}\right]$
\begin{aligned} &=\frac{1}{8} \log \left|\frac{\frac{2 x+3-2}{2}}{\frac{2 x+3+2}{2}}\right|+C \\\\ &\end{aligned}
$=\frac{1}{8} \log \left|\frac{2 x+1}{2 x+5}\right|+C$

Indefinite Integrals Exercise 18.15 Question 2

$\frac{1}{3} \tan ^{-1}\left(\frac{x-5}{3}\right)+C$
Hint:
To solve this problem use special integration formula
Given:
$\int \frac{1}{x^{2}-10 x+34} d x$
Solution:
Let $I=\int \frac{1}{x^{2}-10 x+34} d x$
\begin{aligned} &=\int \frac{1}{x^{2}-2 \cdot x \cdot 5+5^{2}-5^{2}+34} d x \\\\ &=\int \frac{1}{(x-5)^{2}-25+34} d x=\int \frac{1}{(x-5)^{2}+9} d x \\\\ &=\int \frac{1}{(x-5)^{2}+3^{2}} d x \end{aligned}
Put $x-5=t \Rightarrow d x=d t$
Then $I=\int \frac{1}{t^{2}+3^{2}} d t$
$=\frac{1}{3} \tan ^{-1}\left(\frac{t}{3}\right)+C$ $\quad\left[\because \int \frac{1}{x^{2}+a^{2}} d x=\frac{1}{a} \tan ^{-1}\left(\frac{x}{a}\right)+C\right]$
$=\frac{1}{3} \tan ^{-1}\left(\frac{x-5}{3}\right)+C$ $\quad[\because t=x-5]$

Indefinite Integrals Exercise 18.15 Question 3

$\frac{-1}{\sqrt{5}} \log \left|\frac{2 x-1-\sqrt{5}}{2 x-1+\sqrt{5}}\right|+C$
Hint:
To solve this problem use special integration formula
Given:
$\int \frac{1}{1+x-x^{2}} d x$
Solution:
Let $I=\int \frac{1}{1+x-x^{2}} d x=-\int \frac{1}{x^{2}-x-1} d x$
\begin{aligned} &=-\int \frac{1}{x^{2}-2 \cdot x \cdot \frac{1}{2}+\left(\frac{1}{2}\right)^{2}-\left(\frac{1}{2}\right)^{2}-1} d x \\ & \end{aligned}
$=-\int \frac{1}{\left(x-\frac{1}{2}\right)^{2}-\frac{1}{4}-1} d x$
\begin{aligned} &=-\int \frac{1}{\left(x-\frac{1}{2}\right)^{2}-\left(\frac{1+4}{4}\right)} d x \\ & \end{aligned}
$=-\int \frac{1}{\left(x-\frac{1}{2}\right)^{2}-\frac{5}{4}} d x$
Put $x-\frac{1}{2}=t \Rightarrow d x=d t$
Then $I=-\int \frac{1}{t^{2}-\left(\frac{\sqrt{5}}{2}\right)^{2}} d t$
$=\frac{-1}{2 \cdot \frac{\sqrt{5}}{2}} \log \left|\frac{t-\frac{\sqrt{5}}{2}}{t+\frac{\sqrt{5}}{2}}\right|+C \text { }$ $\quad\left[\because \int \frac{1}{x^{2}-a^{2}} d x=\frac{1}{2 a} \log \left|\frac{x-a}{x+a}\right|+C\right]$
$=\frac{-1}{\sqrt{5}} \log \left|\frac{x-\frac{1}{2}-\frac{\sqrt{5}}{2}}{x-\frac{1}{2}+\frac{\sqrt{5}}{2}}\right|+C$ $\quad\left[\because t=x-\frac{1}{2}\right]$
\begin{aligned} &=\frac{-1}{\sqrt{5}} \log \left|\frac{\frac{2 x-1-\sqrt{5}}{2}}{\frac{2 x-1+\sqrt{5}}{2}}\right|+C \\ & \end{aligned}
$=\frac{-1}{\sqrt{5}} \log \left|\frac{2 x-1-\sqrt{5}}{2 x-1+\sqrt{5}}\right|+C$

Indefinite Integrals Exercise 18.15 Question 4

$\frac{1}{3} \log \left|\frac{4 x-4}{4 x+2}\right|+C$
Hint:
To solve this problem use special integration formula
Given:
$\int \frac{1}{2 x^{2}-x-1} d x$
Solution:
Let $I=\int \frac{1}{2 x^{2}-x-1} d x=\frac{1}{2} \int \frac{1}{x^{2}-\frac{x}{2}-\frac{1}{2}} d x$
\begin{aligned} &=\frac{1}{2} \int \frac{1}{x^{2}-2 \cdot x \cdot \frac{1}{4}+\left(\frac{1}{4}\right)^{2}-\left(\frac{1}{4}\right)^{2}-\frac{1}{2}} d x \\ & \end{aligned}
$=\frac{1}{2} \int \frac{1}{\left(x-\frac{1}{4}\right)^{2}-\frac{1}{16}-\frac{1}{2}} d x$
\begin{aligned} &=\frac{1}{2} \int \frac{1}{\left(x-\frac{1}{4}\right)^{2}-\left(\frac{1+8}{16}\right)} d x \\ & \end{aligned}
$=\frac{1}{2} \int \frac{1}{\left(x-\frac{1}{4}\right)^{2}-\frac{9}{16}} d x$
Put $x-\frac{1}{4}=t \Rightarrow d x=d t$
Then $I=\frac{1}{2} \int \frac{1}{t^{2}-\left(\frac{3}{4}\right)^{2}} d x$
$=\frac{1}{2} \cdot \frac{1}{2 \times \frac{3}{4}} \log \left|\frac{t-\frac{3}{4}}{t+\frac{3}{4}}\right|+C$ $\left[\because \int \frac{1}{x^{2}-a^{2}} d x=\frac{1}{2 a} \log \left|\frac{x-a}{x+a}\right|+C\right]$
$=\frac{1}{3} \log \left|\frac{x-\frac{1}{4}-\frac{3}{4}}{x-\frac{1}{4}+\frac{3}{4}}\right|+C \text { }$ $\quad\left[\because t=x-\frac{1}{4}\right]$
\begin{aligned} &=\frac{1}{3} \log \left|\frac{\frac{4 x-1-3}{4}}{\frac{4 x-1+3}{4}}\right|+C \\ & \end{aligned}
$=\frac{1}{3} \log \left|\frac{4 x-4}{4 x+2}\right|+C$

Indefinite Integrals Excercise 18.15 Question 5

$\frac{1}{2} \tan ^{-1}\left(\frac{x+3}{2}\right)+C$
Hint:
To solve this problem use special integration formula
Given:
$\int \frac{1}{x^{2}+6 x+13} d x$
Solution:
Let $I=\int \frac{1}{x^{2}+6 x+13} d x$
\begin{aligned} &=\int \frac{1}{x^{2}+2 \cdot x \cdot 3+3^{2}-3^{2}+13} d x \\ & \end{aligned}
$=\int \frac{1}{(x+3)^{2}-9+13} d x \\$
$=\int \frac{1}{(x+3)^{2}+4} d x$
Put $x+3=t \Rightarrow d x=d t$
Then $I=\int \frac{1}{t^{2}+2^{2}} d t$
$=\frac{1}{2} \tan ^{-1}\left(\frac{t}{2}\right)+C$ $\left[\because \int \frac{1}{x^{2}+a^{2}} d x=\frac{1}{a} \tan ^{-1}\left(\frac{x}{a}\right)+C\right]$
$=\frac{1}{2} \tan ^{-1}\left(\frac{x+3}{2}\right)+C$ $\quad[\because t=x+3]$

Indefinite Integrals Excercise 18.15 Question 6

$\log \left|\frac{(x+3)^{2}}{(x+2)}\right|+C$
Hint:
To solve this problem use special integration formula
Given:
$\int \frac{(x+1)}{(x+2)(x+3)} d x$
Solution:
Let $I=\int \frac{(x+1)}{(x+2)(x+3)} d x$
Applying partial fraction
\begin{aligned} &\frac{(x+1)}{(x+2)(x+3)}=\frac{A}{(x+2)}+\frac{B}{(x+3)} \\ & \end{aligned}
$(x+1)=A(x+3)+B(x+2)$
Putting
\begin{aligned} &x=-2 \\ \end{aligned}
$(-2+1)=A(-2+3)+B(-2+2) \\$
$A=-1$
Putting
\begin{aligned} &x=-3 \\ \end{aligned}
$(-3+1)=A(-3+3)+B(-3+2) \\$
$-2=0+B(-1) \\$
$B=2$
\begin{aligned} &I=\int \frac{x+1}{(x+1)(x+2)} d x=\int \frac{-1}{x+2} d x+\int \frac{2}{x+3} d x \\ \end{aligned}
$=-\log |x+2|+2 \log |x+3| \\$
$=\log \left|\frac{(x+3)^{2}}{(x+2)}\right|+C$

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## RD Sharma Chapter wise Solutions

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The 18th Chapter of RD Sharma class 12th exercise 18.15 has the chapter Indefinite Integrals. The concepts covered under it

• Graphs of indefinite integrals

• concepts of Integrals

• Indefinite integrals of common functions

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