#### explain solution RD Sharma class 12 Chapter 18 Indefinite Integrals exercise 18.30 question 10

$x+\frac{1}{2} \log |x-1|-8 \log |x-2|+\frac{27}{2} \log |x-3|+C$

Hint:

To solve this integration, we use partial fraction method

Given:

$\int \frac{x^{3}}{(x-1)(x-2)(2 x+3)} d x \\$

Explanation:

Let

\begin{aligned} &I=\int \frac{x^{3}}{(x-1)(x-2)(2 x+3)} d x \\ &I=\int \frac{x^{3}}{x^{3}-6 x^{2}+11 x-6} d x \\ &I=\int \frac{\left(x^{3}-6 x^{2}+11 x-6\right)-\left(-6 x^{2}+11 x-6\right)}{x^{3}-6 x+11 x-6} d x \end{aligned}                   [Add and subtract$-6x^{2}+11x+6$]

\begin{aligned} &I=\int\left(1+\frac{6 x^{2}-11 x+6}{x^{3}-6 x+11 x-6}\right) d x \\ &I=\int d x+\int \frac{6 x^{2}-11 x+6}{(x-1)(x-2)(x-3)} d x \\ &I=x+I_{1} \end{aligned}                       (1)

Where

\begin{aligned} &I_{1}=\int \frac{6 x^{2}-11 x+6}{(x-1)(x-2)(x-3)} d x \\ &\frac{6 x^{2}-11 x+6}{(x-1)(x-2)(x-3)}=\frac{A(x-2)(x-3)+B(x-1)(x-3)+C(x-1)(x-2)}{(x-1)(x-2)(x-3)} \\ &6 x^{2}+6-11 x=A(x-2)(x-3)+B(x-1)(x-3)+C(x-1)(x-2) \end{aligned}

\begin{aligned} &\text { At } x=2 \\ &6(4)+6-11(2)=A(0)+B(1)(-1)+C(0) \\ &30-22=-B \\ &8=-B \\ &B=-8 \end{aligned}

\begin{aligned} &\text { At } x=1 \\ &6(1)-11(1)+6=A(-1)(-2)+B(0)+C(0) \\ &1=2 A \\ &A=\frac{1}{2} \end{aligned}

\begin{aligned} &\text { At } x=3 \\ &6(9)-11(3)+6=A(0)+B(0)+C(2)(1) \\ &27=2 C \\ &C=\frac{27}{2} \end{aligned}

\begin{aligned} &\frac{6 x^{2}-11 x+6}{(x-1)(x-2)(x-3)}=\frac{1}{2(x-1)}-\frac{8}{x-2}+\frac{27}{2(x-3)} \\ &I_{1}=\frac{1}{2} \int \frac{1}{x-1} d x-8 \int \frac{1}{x-2} d x+\frac{27}{2} \int \frac{1}{x-3} d x \\ &I_{1}=\frac{1}{2} \log |x-1|-8 \log |x-2|+\frac{27}{2} \log |x-3|+C \end{aligned}

Equation (1)

$I=x+\frac{1}{2} \log |x-1|-8 \log |x-2|+\frac{27}{2} \log |x-3|+C$