#### explain solution RD Sharma class 12 Chapter 18 Indefinite Integrals exercise 18.30 question 24

$\frac{1}{2\left(b^{2}-a^{2}\right)} \log \left|\frac{x-b}{x+a}\right|+\frac{1}{2\left(a^{2}-b^{2}\right)} \log \left|\frac{x-a}{x+b}\right|+C$

Hint:

To solve this integration, we use partial fraction method

Given:

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

Explanation:

Let
\begin{aligned} &I=\int \frac{x}{\left(x^{2}-a^{2}\right)\left(x^{2}-b^{2}\right)} d x \\ &I=\int \frac{x}{(x-a)(x-b)(x+a)(x+b)} d x \ldots\left(x^{2}-y^{2}\right)=(x+y)(x-y) \\ &\frac{x}{(x-a)(x+a)(x-b)(x+b)} d x=\frac{A}{x-a}+\frac{B}{x+a}+\frac{C}{x-b}+\frac{D}{x+b} \end{aligned}

$\! \! \! \! \! \! \! \! \! \! x\! =\! A(x+a)(x-b)(x+b)\! +\! B(x-a)(x-b)(x+b)\! +\! C(x-a)(x+a)(x+b)\! +D\! (x-a)(x-b)(x+a)$                                                                                                                                        (1)

At $x=-a$ equation (1) becomes

\begin{aligned} &-a=A(0)+B(-2 a)(-a-b)(-a+b)+C(0)+(0) \\ &-a=2 B a(a+b)(b-a) \\ &B=\frac{-1}{2\left(b^{2}-a^{2}\right)} \end{aligned}

At $x=a$ equation (1) becomes

\begin{aligned} &a=2 A(a)(a-b)(a+b)+B(0)+C(0)+D(0) \\ &1=2 A\left(a^{2}-b^{2}\right) \\ &A=\frac{1}{2\left(a^{2}-b^{2}\right)} \end{aligned}

At $x=b$ equation (1) becomes

\begin{aligned} &b=A(0)+B(0)+C(2 b)(b-a)(b+a) \\ &b=2 b C\left(b^{2}-a^{2}\right) \\ &C=\frac{1}{2\left(b^{2}-a^{2}\right)} \end{aligned}

At $x=-b$equation (1) becomes

\begin{aligned} &-b=A(0)+B(0)+C(0)+(-2 b)(-b-a)(-b+a) \\ &-b=2 D b(b+a)(a-b) \\ &\, D=\frac{-1}{a^{2}-b^{2}} \end{aligned}

\begin{aligned} \frac{x}{(x-a)(x+a)(x-b)(x+b)} &=\frac{1}{2\left(a^{2}-b^{2}\right)(x-a)}+\frac{(-1)}{2\left(b^{2}-a^{2}\right)(x+a)}+\frac{1}{2\left(b^{2}-a^{2}\right)(x-b)}-\frac{1}{2\left(a^{2}-b^{2}\right)(x+b)} \\ \end{aligned}

\begin{aligned} &I=\frac{1}{2\left(a^{2}-b^{2}\right)} \int \frac{1}{x-a} d x+\frac{1}{2\left(b^{2}-a^{2}\right)} \int \frac{d x}{x-b}-\frac{1}{2\left(b^{2}-a^{2}\right)} \int \frac{d x}{x+a}-\frac{1}{2\left(a^{2}-b^{2}\right)} \int \frac{d x}{x+b} \\ &I=\frac{1}{2\left(a^{2}-b^{2}\right)} \log |x-a|+\frac{1}{2\left(b^{2}-a^{2}\right)} \log |x-b|-\frac{1}{2\left(b^{2}-a^{2}\right)} \log |x+a| \end{aligned}

$\begin{gathered} I=\frac{1}{2\left(a^{2}-b^{2}\right)} \log |x-a|+\frac{1}{2\left(b^{2}-a^{2}\right)} \log |x-b|-\frac{1}{2\left(b^{2}-a^{2}\right)} \log |x+a| -\frac{1}{2\left(a^{2}-b^{2}\right)} \log |x+b|+C\\ \end{gathered}$

$I=\frac{1}{2\left(a^{2}-b^{2}\right)} \log \left|\frac{x-a}{x+b}\right|+\frac{1}{2\left(b^{2}-a^{2}\right)} \log \left|\frac{x-b}{x+a}\right|+C$