explain solution RD Sharma class 12 Chapter 18 Indefinite Integrals exercise 18.30 question 41

$\frac{1}{3} \tan ^{-1} x-\frac{1}{6} \tan ^{-1} \frac{x}{2}+C$

Hint:

To solve this integration, we use partial fraction method

Given:

$\int \frac{1}{\left(x^{2}+1\right)\left(x^{2}+4\right)} d x \\$

Explanation:

Let

\begin{aligned} &I=\int \frac{1}{\left(x^{2}+1\right)\left(x^{2}+4\right)} d x \\ &\frac{1}{\left(x^{2}+1\right)\left(x^{2}+4\right)}=\frac{A x+B}{x^{2}+1}+\frac{C x+D}{x^{2}+4} \\ &1=(A x+B)\left(x^{2}+4\right)+(C x+D)\left(x^{2}+1\right) \\ &1=x^{3}(A+C)+x^{2}(B+D)+(4 A+C) x+4 B+D \end{aligned}

Equating the similar terms

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

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

\begin{aligned} &4 A+C=0 \\ &-4 C+C=0 \end{aligned}                [From the equation (1)]

\begin{aligned} &-3 C=0 \\ &C=0 \\ &A=0 \end{aligned}                          [From the equation (1)]

\begin{aligned} &4 B+D=1 \\ &-4 D+D=1 \\ &-3 D=1 \end{aligned}

$D=\frac{-1}{3}$                               (3)

$B=\frac{1}{3}$                                  [From the equation (2)]

\begin{aligned} &\frac{1}{\left(x^{2}+1\right)\left(x^{2}+4\right)}=\frac{1}{3\left(x^{2}+1\right)}+\frac{(-1)}{3\left(x^{2}+4\right)} \\ &I=\frac{1}{3} \int \frac{1}{x^{2}+1} d x-\frac{1}{3} \int \frac{1}{x^{2}+4} d x \\ &I=\frac{1}{3} \tan ^{-1} x-\frac{1}{3} \cdot\left(\frac{1}{2}\right) \tan ^{-1} \frac{x}{2}+C \\ &I=\frac{1}{3} \tan ^{-1} x-\frac{1}{6} \tan ^{-1} \frac{x}{2}+C \end{aligned}