#### explain solution RD Sharma class 12 Chapter 18 Indefinite Integrals exercise 18.30 question 42

$\frac{2}{\sqrt{3}} \tan ^{-1}\left(\frac{\sqrt{3} x}{2}\right)-\tan ^{-1} x+C$

Hint:

To solve this integration, we use partial fraction method

Given:

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

Explanation:

Let

\begin{aligned} &I=\int \frac{x^{2}}{\left(x^{2}+1\right)\left(3 x^{2}+4\right)} d x \\ &\frac{x^{2}}{\left(x^{2}+1\right)\left(3 x^{2}+4\right)}=\frac{A x+B}{x^{2}+1}+\frac{C x+D}{3 x^{2}+4} \\ &x^{2}=(A x+B)\left(3 x^{2}+4\right)+(C x+D)\left(x^{2}+1\right) \\ &x^{2}=(3 A+C) x^{3}+(3 B+D) x^{2}+(4 A+C) x+(4 B+D) \end{aligned}

Equating the similar term

\begin{aligned} &3 A+C=0 \\ &3 A=-C \end{aligned}                            (1)

$4A+C= 0$

$4A-3A= 0$                   [From equation (1)]

$A= 0$                               (2)

Equation (1)

\begin{aligned} &C=0 \\ &3 B+D=1 \end{aligned}                      (3)

\begin{aligned} &4 B+D=0 \\ &D=-4 B \end{aligned}                     (4)

Equation (3)

\begin{aligned} &3 B-4 B=1 \\ &-B=1 \\ &B=-1 \\ &D=4 \end{aligned}                               [From equation (4)]

\begin{aligned} &I=-1 \int \frac{1}{x^{2}+1} d x+4 \int \frac{1}{3 x^{2}+4} d x \\ &I=-\tan ^{-1} x+\frac{4}{3} \int \frac{1}{x^{2}+\frac{4}{3}} d x+C \end{aligned}
\begin{aligned} &I=-\tan ^{-1} x+\frac{4}{3} \int \frac{1}{x^{2}+\left(\frac{2}{\sqrt{3}}\right)^{2}} d x+C \\ &I=-\tan ^{-1} x+\frac{4}{3}\left(\frac{\sqrt{3}}{2}\right) \tan ^{-1}\left(\frac{x}{\frac{2}{\sqrt{3}}}\right)+C \\ &I=-\tan ^{-1} x+\frac{2}{\sqrt{3}} \tan ^{-1}\left(\frac{\sqrt{3} x}{2}\right)+C \end{aligned}