#### Need solution for RD Sharma Maths Class 12 Chapter 18 Indefinite Integrals Excercise 18.30 Question 35

$\frac{9}{4} \log |x+2|-\frac{9}{8} \log \left|x^{2}+4\right|+\frac{9}{4} \tan ^{-1} \frac{x}{2}+C$

Hint:

To solve this integration, we use partial fraction method

Given:

$\int \frac{18}{(x+2)\left(x^{2}+4\right)} d x \\$

Explanation:

Let

\begin{aligned} &I=\int \frac{18}{(x+2)\left(x^{2}+4\right)} d x \\ &\frac{18}{(x+2)\left(x^{2}+4\right)}=\frac{A}{x+2}+\frac{B x+C}{x^{2}+4} \\ &18=A\left(x^{2}+4\right)+(B x+C)(x+2) \\ &18=x^{2}(A+B)+(2 B+C) x+(4 A+2 C) \end{aligned}

Equating the similar terms

\begin{aligned} &A+B=0 \\ &A=-B \end{aligned}                            (1)

\begin{aligned} &2 B+C=0 \\ &C=-2 B \end{aligned}                            (2)

\begin{aligned} &4 A+2 C=18 \\ &-4 B-4 B=18 \\ &-8 B=18 \\ &B=\frac{-9}{4} \end{aligned}

Equation (2)

\begin{aligned} &C=-2 \times\left(\frac{-9}{4}\right) \\ &C=\frac{9}{2} \end{aligned}

Equation (1)

\begin{aligned} A=\frac{9}{4} \\ \end{aligned}

\begin{aligned} &\frac{18}{(x+2)\left(x^{2}+4\right)}=\frac{9}{4(x+2)}+\frac{\frac{-9}{4} x+\frac{9}{2}}{x^{2}+4} \\ &=\frac{9}{4(x+2)}+\frac{-9 x+18}{4\left(x^{2}+4\right)} \\ &I=\frac{9}{4} \int \frac{d x}{x+2}-\frac{9}{8} \int \frac{2 x}{x^{2}+4} d x+\frac{9}{2} \int \frac{1}{x^{2}+4} d x \end{aligned}

$\frac{9}{4} \log |x+2|-\frac{9}{8} \log \left|x^{2}+4\right|+\frac{9}{4} \tan ^{-1} \frac{x}{2}+C$