#### explain solution RD Sharma class 12 Chapter 18 Indefinite Integrals exercise 18.30 question 27

$\frac{11}{4} \log \left|\frac{x+1}{x+3}\right|+\frac{5}{2(x+1)}+C$

Hint:

To solve this integration, we use partial fraction method

Given:

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

Explanation:

Let

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

\begin{aligned} &3 x-2=A\left(x^{2}+4 x+3\right)+B(x+3)+C\left(x^{2}+1+2 x\right) \\ &3 x-2=(A+C) x^{2}+x(4 A+B+2 C)+(3 A+3 B+C) \end{aligned}

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

$A+C=0$

$A=-C$                               (1)

$4A+B+2C=3$             (2)

$-4C+B+2C=3$          from equation (1)

$B-2C=3$                        (3)

\begin{aligned} &3 A+3 B+C=-2 \\ &-3 C+3 B+C=-2 \end{aligned}       from equation (1)

$3B-2C=-2$                  (4)

Subtract equation (3) from equation (4)

\begin{aligned} &3 B-2 C=-2- \\ &B-2 C=3 \\ &\overline{ 2 B=-5} \\ &B=\frac{-5}{2} \end{aligned}

Equation (3)

\begin{aligned} &\frac{-5}{2}-2 C=3 \\ &\frac{-5}{2}-3=2 C \\ &\frac{-11}{2}=2 C \\ &C=\frac{-11}{4} \\ &A=\frac{11}{4} \end{aligned}
$\frac{3 x-2}{(x+1)^{2}(x+3)}=\frac{11}{4(x+1)}-\frac{5}{2(x+1)^{2}}-\frac{11}{4(x+3)}$

Thus

\begin{aligned} &I=\frac{11}{4} \int \frac{1}{x+1} d x-\frac{5}{2} \int \frac{1}{(x+1)^{2}} d x-\frac{11}{4} \int \frac{1}{x+3} d x \\ &I=\frac{11}{4} \log |x+1|-\frac{5}{2}\left(\frac{1}{-1(x+1)}\right)-\frac{11}{4} \log |x+3|+C \\ &I=\frac{11}{4} \log \left|\frac{x+1}{x+3}\right|+\frac{5}{2(x+1)}+C \end{aligned}