#### explain solution RD Sharma class 12 Chapter 18 Indefinite Integrals exercise 18.30 question 2

$\frac{1}{8} \log \left|\frac{x(x-4)}{(x-2)^{2}}\right|+C$

Hint:

To solve the given integration, we use partial fraction method

Given:

$\int \frac{1}{x(x-2)(x-4)} d x$

Explanation:

Let $I=\int \frac{1}{x(x-2)(x-4)} d x$

Now express the functions in terms of partial fraction

$\frac{1}{x(x-2)(x-4)}=\frac{A}{x}+\frac{B}{x-2}+\frac{C}{x-4} \quad\left[\because \frac{N(x)}{(a x-b)(c x-d)}=\frac{A}{a x-b}+\frac{B}{c x-d}\right]$

\begin{aligned} &\frac{1}{x(x-2)(x-4)}=\frac{A(x-2)(x-4)+B x(x-4)+C x(x-2)}{x(x-2)(x-4)} \\ &1=A\left(x^{2}-6 x+8\right)+B\left(x^{2}-4 x\right)+C\left(x^{2}-2 x\right) \\ &1=x^{2}(A+B+C)+x(-6 A-4 B-2 C)+8 A \end{aligned}

On comparing coefficient, we get

\begin{aligned} &A+B+C=0\quad\quad\quad\quad(1)\\ &-6 A-4 B-2 C=0\quad\quad\quad(2)\\ &8 A=1 \quad\quad\quad\quad(3)\\ &A=\frac{1}{8} \end{aligned}

Now equation (2)

\begin{aligned} &\Rightarrow \frac{-6}{8}-4 B-2 C=0 \\ &\Rightarrow 2 B+C=\frac{-3}{8} \end{aligned}  (4)

Equation (1)

\begin{aligned} &\Rightarrow \frac{1}{8}+B+C=0 \\ &\Rightarrow B+C=\frac{-1}{8} \end{aligned}     (5)

Subtract equation (5) from equation (4)

$\begin{gathered} 2 B+C=\frac{-3}{8}- \\ B+C=\frac{-1}{8} \\ \hline B=\frac{-2}{8} \end{gathered}$

Now equation (5)

\begin{aligned} &\Rightarrow \frac{-2}{8}+C=\frac{-1}{8} \\ &C=\frac{-1}{8}+\frac{2}{8} \\ &C=\frac{1}{8} \end{aligned}

Now

\begin{aligned} &\frac{1}{x(x-2)(x-4)}=\left(\frac{1}{8 x}\right)+\frac{(-2)}{8(x-2)}+\frac{1}{8(x-4)} \\ &I=\int\left(\frac{1}{8 x}-\frac{2}{8(x-2)}+\frac{1}{8(x-4)}\right) d x \\ &I=\frac{1}{8} \int \frac{1}{x} d x-\frac{2}{8} \int \frac{1}{x-2} d x+\frac{1}{8} \int \frac{d x}{x-4} \end{aligned}

\begin{aligned} &I=\frac{1}{8} \log |x|-\frac{2}{8} \log |x-2|+\frac{1}{8} \log |x-4|+C \quad\left[\int \frac{1}{a x-b} d x=\frac{1}{a} \log |a x-b|\right] \\ &I=\frac{1}{8} \log \left|\frac{x(x-4)}{(x-2)^{2}}\right|+C \end{aligned}