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

$\frac{-2}{5} \tan ^{-1} \frac{x}{2}+\frac{3}{5} \tan \frac{x}{3}+C$

Hint:

To solve this integration, we use partial fraction method

Given:

$\int \frac{x^{2}}{\left(x^{2}+4\right)\left(x^{2}+9\right)} d x \\$

Explanation:

Let

\begin{aligned} &I=\int \frac{x^{2}}{\left(x^{2}+4\right)\left(x^{2}+9\right)} d x \\ &x^{2}=(A+B x)\left(x^{2}+9\right)+(C x+D)\left(x^{2}+4\right) \\ &x^{2}=x^{3}(B+C)+x^{2}(A+D)+x(9 B+4 C)+(9 A+4 D) \end{aligned}

Comparing the coefficient

$B+C=0$               (1)

$B=-C$

$A+D=1$                 (2)

$9B+4C=0$           (3)

$9A+4D=1$           (4)

Equation (3)

\begin{aligned} &9(-C)+4 C=0 \\ &-9 C+4 C=0 \\ &5 C=0 \\ &C=0 \\ &B=0 \end{aligned}

Equation (2)

$A=1-D$

Equation (4)

\begin{aligned} &9(1-D)+4 D=0 \\ &9-9 D+4 D=0 \\ &9=5 D \\ &D=\frac{9}{5} \\ &A=1-\frac{9}{5} \\ &A=\frac{-4}{5} \end{aligned}

Now

\begin{aligned} &\frac{x^{2}}{\left(x^{2}+4\right)\left(x^{2}+9\right)}=\frac{-4}{5\left(x^{2}+4\right)}+\frac{9}{5\left(x^{2}+9\right)} \\ &I=\frac{-4}{5} \int \frac{1}{x^{2}+4} d x+\frac{9}{5} \int \frac{1}{x^{2}+9} d x \end{aligned}

\begin{aligned} &I=\frac{-4}{5}\left(\frac{1}{2}\right) \tan ^{-1} \frac{x}{2}+\frac{9}{5}\left(\frac{1}{3}\right) \tan ^{-1} \frac{x}{3}+C \\ &I=\frac{-2}{5} \tan ^{-1} \frac{x}{2}+\frac{3}{5} \tan ^{-1} \frac{x}{3}+C \end{aligned}