# Get Answers to all your Questions

#### explain solution RD Sharma class 12 Chapter 18 Indefinite Integrals exercise 18.30 question 20

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

Hint:

To solve this integration, we use partial fraction method

Given:

$\int \frac{x^{2}+6 x-8}{x^{3}-4 x} d x \\$

Explanation:

Let

\begin{aligned} &I=\int \frac{x^{2}+6 x-8}{x^{3}-4 x} d x \\ &I=\int \frac{x^{2}+6 x-8}{x\left(x^{2}-4\right)} d x \end{aligned}

\begin{aligned} &I=\int \frac{x^{2}+6 x-8}{x(x-2)(x+2)} d x \ldots .\left[x^{2}-y^{2}=(x+y)(x-y)\right] \\ &\frac{x^{2}+6 x-8}{x(x-2)(x+2)}=\frac{A}{x}+\frac{B}{x-2}+\frac{C}{x+2} \\ &x^{2}+6 x-8=A(x-2)(x+2)+B x(x+2)+C x(x-2) \end{aligned}

\begin{aligned} &\text { At } x=0 \\ &0+6(0)-8=A(-2)(2)+0+0 \\ &-8=-4 A \\ &A=2 \end{aligned}

\begin{aligned} &\text { At } x=2 \\ &4+6(2)-8=B(2)(4) \\ &8=8 B \\ &B=1 \end{aligned}

\begin{aligned} &\text { At } x=-2 \\ &4+6(-2)-8=A(0)+B(0)+C(-2)(-4) \\ &-16=8 C \\ &C=-2 \end{aligned}

Now

\begin{aligned} &\frac{x^{2}+6 x-8}{x(x-2)(x+2)}=\frac{2}{x}+\frac{1}{x-2}-\frac{2}{x+2} \\ &I=2 \int \frac{1}{x} d x+\int \frac{1}{x-2} d x-2 \int \frac{1}{x+2} d x \\ &I=2 \log |x|+\log |x-2|-2 \log |x+2|+C \\ &I=\log \left|\frac{x^{2}(x-2)}{(x+2)^{2}}\right|+C \end{aligned}