#### explain solution RD Sharma class 12 Chapter 18 Indefinite Integrals exercise 18.30 question 28

$\frac{3}{25} \log \left|\frac{x-3}{x+2}\right|-\frac{7}{5(x-3)}+C$

Hint:

To solve this integration, we use partial fraction method

Given:

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

Explanation:

Let

\begin{aligned} &I=\int \frac{2 x+1}{(x+2)(x-3)^{2}} d x \\ &\frac{2 x+1}{(x+2)(x-3)^{2}}=\frac{A}{x+2}+\frac{B}{x-3}+\frac{C}{(x-3)^{2}} \\ &2 x+1=A(x-3)^{2}+B(x-3)(x+2)+C(x+2) \\ &2 x+1=x^{2}(A+B)+x(-6 A-B+C)+(9 A-6 B+2 C) \end{aligned}

Comparing the coefficient of $x^{2},x$and constant term

$A+B=0$                    (1)

\begin{aligned} &A=-B \\ &-6 A-B+C=2 \\ &-6 A+A+C=2 \end{aligned}           [From equation 1]

$-5A+C=2$                   (2)

\begin{aligned} &9 A-6 B+2 C=1 \\ &9 A+6 A+2 C=1 \\ &15 A+2 C=1 \end{aligned}                   (3)

Multiply the equation (2) by 2 and subtract it from equation (3)

\begin{aligned} &15 A+2 C=1\\ &\frac{-10 A+2 C=4}{25 A=-3}\\ &A=\frac{-3}{25}\\ &B=\frac{3}{25} \end{aligned}                               [From the equation (1)]

Equation (2)

\begin{aligned} &-5\left(\frac{-3}{25}\right)+C=2 \\ &\frac{3}{5}+C=2 \\ &x=2-\frac{3}{5} \\ &C=\frac{7}{5} \end{aligned}

\begin{aligned} &\frac{2 x+1}{(x+2)(x-3)^{2}}=\frac{-3}{25(x+2)}+\frac{3}{25(x-3)}+\frac{7}{5(x-3)^{2}} \\ &I=\frac{-3}{25} \int \frac{1}{x+2} d x+\frac{3}{25} \int \frac{1}{x-3}+\frac{7}{5} \int \frac{1}{(x-3)^{2}} \\ &I=\frac{-3}{25} \log |x+2|+\frac{3}{25} \log |x-3|+\frac{7}{5} \cdot \frac{1}{(-1)(x-3)}+C \\ &I=\frac{3}{25} \log \left|\frac{x-3}{x+2}\right|-\frac{7}{5(x-3)}+C \end{aligned}