#### explain solution RD Sharma class 12 Chapter 18 Indefinite Integrals exercise 18.30 question 22

$\log \left|\frac{(2 \log x+1)^{\square}}{(3 \log x+2)^{1 / 2}}\right|+C$

Hint:

To solve this integration, we use partial fraction method

Given:

$\int \frac{1}{x\left[6(\log x)^{2}+7 \log x+2\right]} d x$

Explanation:

$I=\int \frac{1}{x\left[6(\log x)^{2}+7 \log x+2\right]} d x$

Let

\begin{aligned} &\log x=y \\ &\frac{1}{x} d x=d y \\ &I=\int \frac{d y}{6 y^{2}+7 y+2} d x \\ &I=\int \frac{d y}{(3 y+2)(2 y+1)} \end{aligned}

\begin{aligned} &\frac{1}{(3 y+2)(2 y+1)}=\frac{A}{3 y+2}+\frac{B}{2 y+1} \\ &1=A(2 y+1)+B(3 y+2) \\ &1=y(2 A+3 B)+(A+2 B) \end{aligned}

Comparing the coefficient

$2A+3B=0$                (1)

$A+2B=1$                   (2)

Multiply equation (2) by 2 and then

Subtract equation (1) from it

\begin{aligned} &2 A+4 B=2 \quad- \\ &\frac{2 A+3 B=0}{B=2} \end{aligned}

Equation (2)

\begin{aligned} &A+2(2)=1 \\ &A+4=1 \\ &A=-3 \end{aligned}

Now

\begin{aligned} &\frac{1}{(3 y+2)(2 y+1)}=\frac{-3}{3 y+2}+\frac{2}{2 y+1} \\ &I=-3 \int \frac{1}{3 y+2} d y+2 \int \frac{1}{2 y+1} d y \\ &I=-3 \cdot \frac{1}{3} \log |3 y+2|+2 \cdot \frac{1}{2} \log |2 y+1| \\ &I=\log\left | \frac{2y+1}{3y+2} \right |+C \end{aligned}

As $y=\log x$

$I=\log \left|\frac{(2 \log x+1)^{\square}}{(3 \log x+2)}\right|+C$