#### explain solution RD Sharma class 12 Chapter 18 Indefinite Integrals exercise 18.30 question 13

$\frac{1}{2} \log \left|\frac{\log x}{\log x+2}\right|+C$

Hint:

To solve this integration, we use partial fraction method

Given:

$\int \frac{1}{x \log x(2+\log x)} d x$

Explanation:

$I=\int \frac{1}{x \log x(2+\log x)} d x$

Let

\begin{aligned} &\log x=y \\ &\frac{1}{x}=d y \\ &I=\int \frac{d y}{y(y+2)} \end{aligned}

Now

\begin{aligned} &\frac{1}{y(y+2)}=\frac{A}{y}+\frac{B}{y+2} \\ &\frac{1}{y(y+2)}=\frac{A(y+2)+B y}{(y+2) y} \\ &1=A(2+y)+B y \\ &1=2 A+(A+B) y \end{aligned}

Comparing the coefficient

\begin{aligned} &2 A=1 \\ &A=\frac{1}{2} \\ &A+B=0 \\ &B=-A \\ &B=-A \\ &B=\frac{-1}{2} \end{aligned}

\begin{aligned} &\frac{1}{y(y+2)}=\frac{1}{2 y}-\frac{1}{2(y+2)} \\ &I=\frac{1}{2} \int \frac{1}{y} d y-\frac{1}{2} \int \frac{d y}{y+2} \end{aligned}

\begin{aligned} &I=\frac{1}{2} \log |y|-\frac{1}{2} \log |y+2|+C \\ &I=\frac{1}{2} \log \left|\frac{y}{y+2}\right|+C \\ &I=\frac{1}{2} \log \left|\frac{\log x}{\log x+2}\right|+C \end{aligned}