#### Need Solution for R.D.Sharma Maths Class 12 Chapter 18 Indefinite Integrals Exercise 18.19 Question 7  Maths Textbook Solution.

Answer:$-\frac{1}{2} \ln \left(3 x^{2}+4 x+2\right)+\frac{3}{\sqrt{2}} \tan ^{-1}\left(\frac{3 x+2}{\sqrt{2}}\right)+c$

Hint:Use $\int \frac{P n+q}{a x^{2}+b x+c}$

Given: $\int \frac{1-3 x}{3 x^{2}+4 x+2} d x$

Solution:

$I=\int \frac{1-3 x}{3 x^{2}+4 x+2} d n$

$\int \frac{P n+q}{a x^{2}+b x+c} d x$

$\text { Numerator }=A\left(\frac{d}{d x}(\text { deno min ator })\right)+B$

$1-3 x=A\left(\frac{d}{d x}\left(3 x^{2}+4 x+2\right)\right)+B$

$1-3 x=A(6 x+4)+B$

$1-3 x=6 A x+4 A+B$

On comparing,

$6 A=-3 \Rightarrow A=-\frac{3}{6} \Rightarrow A=\frac{-1}{2}$

$4 A+B=1 \Rightarrow 4\left(\frac{-1}{2}\right)+B=1 \Rightarrow-2+B=1 \Rightarrow B=3$

$I=\int \frac{\left(-\frac{1}{2}\right)(6 x+4)+3}{3 x^{2}+4 x+2} d x$

$I=\int-\frac{1}{2} \frac{(6 x+4)}{3 x^{2}+4 x+2} d x+\int \frac{3}{3 x^{2}+4 x+2} d x$

$=\frac{-1}{2} \int \frac{(6 x+4)}{3 x^{2}+4 x+2} d x+\int \frac{3}{3\left(x^{2}+\frac{4}{3} x+\frac{2}{3}\right)} d x$

$=\frac{-1}{2} \int \frac{(6 x+4)}{3 x^{2}+4 x+2} d x+\int \frac{1}{x^{2}+\frac{4}{3} x+\frac{4}{9}-\frac{4}{9}+\frac{2}{3}} d x$

$=\frac{-1}{2} \int \frac{\frac{d}{d x}\left(3 x^{2}+4 x+2\right)}{3 x^{2}+4 x+2} d x+\int \frac{1}{\left(x+\frac{2}{3}\right)^{2}+\left(\frac{\sqrt{2}}{3}\right)^{2}} d x$

\begin{aligned} &3 x^{2}+4 x+2=t \\ &\frac{d\left(3 x^{2}+4 x+2\right)}{d x}=d t \end{aligned}

$=-\frac{1}{2} \int \frac{d t}{t}+\frac{1}{\frac{\sqrt{2}}{3}}\left[\frac{\tan ^{-1} x+\frac{2}{3}}{\frac{\sqrt{2}}{3}}\right]\left[\int \frac{1}{x^{2}+a^{2}}=\frac{1}{a} \tan ^{-1} \frac{x}{2}+c\right]$

$=-\frac{1}{2} \ln \left(3 x^{2}+4 x+2\right)+\frac{3}{\sqrt{2}} \tan ^{-1}\left(\frac{3 x+2}{\sqrt{2}}\right)+c$