#### explain solution RD Sharma class 12 Chapter 18 Indefinite Integrals exercise 18.30 question 58

$\frac{a}{a^{2}-b^{2}} \tan ^{-1} \frac{x}{a}+\frac{b}{b^{2}-a^{2}} \tan ^{-1} \frac{x}{b}+C$

Hint:

To solve this integration, we use partial fraction method

Given:

$\int \frac{x^{2}}{\left(x^{2}+a^{2}\right)\left(x^{2}+b^{2}\right)} d x$

Explanation:

Let

$I=\int \frac{x^{2}}{\left(x^{2}+a^{2}\right)\left(x^{2}+b^{2}\right)} d x$

$\frac{x^{2}}{\left(x^{2}+a^{2}\right)\left(x^{2}+b^{2}\right)}=\frac{A x+B}{\left(x^{2}+a^{2}\right)}+\frac{C x+D}{\left(x^{2}+b^{2}\right)} \\$        (i)

$x^{2}=(A x+B)\left(x^{2}+b^{2}\right)+(C x+D)\left(x^{2}+a^{2}\right)$

On comparing coefficient

$x^{2}$Coefficient

$1=B+D$         (1)

$x$ Coefficient

$0=A b^{2}+C a^{2}$               (2)

Constant

$0=B b^{2}+D a^{2}$              (3)

\begin{aligned} &x^{3} \text { Coefficient }\\ &0=A+C \end{aligned}                       (4)

Solving (2) and (4)

\begin{aligned} &A b^{2}+C a^{2}=0 \\ &A+C=0 \\ &C=-A \\ &A b^{2}-A a^{2}=0 \\ &A\left(b^{2}-a^{2}\right)=0 \\ &A=0, C=0 \end{aligned}

$B+D=1$                    (1)

$B b^{2}+D a^{2}=0$              (3)

Multiply (1) by $b^{2}$and subtract (3)

\begin{aligned} &B b^{2}+D b^{2}=b^{2}- \\ &B b^{2}+D a^{2}=0 \\ &\overline{D\left(b^{2}-a^{2}\right)=b^{2}} \end{aligned}

\begin{aligned} &D=\frac{b^{2}}{b^{2}-a^{2}} \\ &B=1-D \\ &B=1-\frac{b^{2}}{b^{2}-a^{2}} \\ &B=\frac{-a^{2}}{b^{2}-a^{2}} \end{aligned}

(i) Becomes
\begin{aligned} &\frac{x^{2}}{\left(x^{2}+a^{2}\right)\left(x^{2}+b^{2}\right)} d x=\frac{-a^{2}}{b^{2}-a^{2}} \int \frac{1}{x^{2}+a^{2}} d x+\frac{b^{2}}{b^{2}-a^{2}} \int \frac{1}{x^{2}+b^{2}} d x \\ &=\frac{a^{2}}{a^{2}-b^{2}} \cdot \frac{1}{a} \tan ^{-1} \frac{x}{a}+\frac{b^{2}}{b^{2}-a^{2}} \cdot \frac{1}{b} \tan ^{-1} \frac{x}{b} \\ &=\frac{a}{a^{2}-b^{2}} \tan ^{-1} \frac{x}{a}+\frac{b}{b^{2}-a^{2}} \tan ^{-1} \frac{x}{b}+C \end{aligned}