#### explain solution RD Sharma class 12 Chapter 18 Indefinite Integrals exercise 18.30 question 29

$\frac{3}{5} \log |x-2|+\frac{2}{5} \log |x+3|-\frac{1}{x-2}+C$

Hint:

To solve this integration, we use partial fraction method

Given:

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

Explanation:

Let

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

\begin{aligned} &x^{2}+1=A(x-2)(x+3)+B(x+3)+C(x-2)^{2} \\ &x^{2}+1=A\left(x^{2}+x-6\right)+B(x+3)+C\left(x^{2}+4-4 x\right) \\ &x^{2}+1=(A+C) x^{2}+(A+B-4 C) x+(-6 A+3 B+4 C) \end{aligned}

Equating the similar terms, we get

$A+C=1$                        (1)

$A+B-4C=0$                (2)

$-6A+3B+4C=1$        (3)

Subtract equation (1) from equation (2) and we get

$B-5C=-1$                     (4)

Multiply equation (1) by 6 and then adding equation (3)

$3B+10C=7$                   (5)

Multiply equation (4) by 3 and then subtract it from equation (5)

$\! \! \! \! \! \! \! \! 3 B+10 C=7 \\ 3 B-15 C=-3 \\ \overline{25 C=10 }\\\\$

\begin{aligned} &C=\frac{10}{25} \\ &C=\frac{2}{5} \end{aligned}

Equation (4)

\begin{aligned} &B-5\left(\frac{2}{5}\right)=-1 \\ &B-2=-1 \\ &B=1 \end{aligned}

Equation (1)

\begin{aligned} &A+\frac{2}{5}=1 \\ &A=1-\frac{2}{5} \\ &A=\frac{3}{5} \end{aligned}

Now

$\frac{x^{2}+1}{(x-2)^{2}(x+3)}=\frac{3}{5(x-2)}+\frac{1}{(x-2)^{2}}+\frac{2}{5(x+3)}$

Thus

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