#### explain solution RD Sharma class 12 Chapter 18 Indefinite Integrals exercise 18.30 question 17

$\frac{1}{6} \log |x-1|-\frac{1}{2} \log |x+1|+\frac{1}{3} \log |x+2|+C$

Hint:

To solve this integration, we use partial fraction method

Given:

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

Explanation:

Let

\begin{aligned} &I=\int \frac{1}{(x-1)(x+1)(x+2)} d x \\ &\frac{1}{(x-1)(x+2)(x+1)}=\frac{A}{x-1}+\frac{B}{x+1}+\frac{C}{x+2} \end{aligned}

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

\begin{aligned} &\text { At } x=1 \\ &\begin{array}{l} 1=A(2)(3)+B(0)+C(0) \\ 1=6 A \\ A=\frac{1}{6} \end{array} \end{aligned}
\begin{aligned} &\text { At } x=-1 \\ &1=A(0)+B(-2)(1)+C(0) \\ &1=-2 B \\ &B=\frac{-1}{2} \end{aligned}
\begin{aligned} &\text { At } x=-2 \\ &1=A(0)+B(0)+C(-3)(-1) \\ &1=3 C \\ &C=\frac{1}{3} \end{aligned}
\begin{aligned} &\frac{1}{(x-1)(x+1)(x+2)}=\frac{1}{6(x-1)}-\frac{1}{2(x+1)}+\frac{1}{3(x+2)} \\ &I=\frac{1}{6} \int \frac{1}{x-1} d x-\frac{1}{2} \int \frac{1}{x+1} d x+\frac{1}{3} \int \frac{1}{x+2} d x \\ &I=\frac{1}{6} \log |x-1|-\frac{1}{2} \log |x+1|+\frac{1}{3} \log |x+2|+C \end{aligned}