#### explain solution RD Sharma class 12 Chapter 18 Indefinite Integrals exercise 18.30 question 25

$\frac{-1}{14} \tan ^{-1} \frac{x}{2}+\frac{8}{35} \tan ^{-1} \frac{x}{5}+C$

Hint:

To solve this integration, we use partial fraction method

Given:

$\int \frac{x^{2}+1}{\left(x^{2}+4\right)\left(x^{2}+25\right)} d x \\$

Explanation:

Let

\begin{aligned} &I=\int \frac{x^{2}+1}{\left(x^{2}+4\right)\left(x^{2}+25\right)} d x \\ &\frac{x^{2}+1}{\left(x^{2}+4\right)\left(x^{2}+25\right)}=\frac{A x+B}{x^{2}+4}+\frac{C x+D}{x^{2}+25} \\ &x^{2}+1=(A x+B)\left(x^{2}+25\right)+(C x+D)\left(x^{2}+4\right) \\ &x^{2}+1=x^{3}(A+C)+x^{2}(B+D)+x(25 A+4 C)+(25 B+4 D) \end{aligned}

Comparing the coefficient

Coefficient of $x^{3}$

$A+C=0$                    (2)

$A=-C$                       (3)

Coefficient of $x^{2}$

$B+D=1$                    (4)

Coefficient of $x$

$25A+4C=0$

$-21C+4C=0$                               [From the equation (3)]

$-21C=0$

$C=0$                                   (5)

$A=0$                                   (6)

Constant term

$25 B+4 D=1$                   (7)

Multiply the equation (4) by 4 and then subtract it from equation (7)

\begin{aligned} &25 B+4 D=1- \\ &4 B+4 D=4 \\ &\overline{21 B=-3} \\ &B=\frac{-1}{7} \end{aligned}

Equation (4)

\begin{aligned} &\frac{-1}{7}+D=1 \\ &D=1+\frac{1}{7} \\ &D=\frac{8}{7} \\ &\frac{\left(x^{2}+1\right)}{\left(x^{2}+4\right)\left(x^{2}+25\right)}=\frac{-1}{7\left(x^{2}+4\right)}+\frac{8}{7\left(x^{2}+25\right)} \end{aligned}

Thus

\begin{aligned} &I=\frac{-1}{7} \int \frac{1}{x^{2}+4} d x+\frac{8}{7} \int \frac{1}{x^{2}+25} d x \\ &I=\frac{-1}{7} \cdot\left(\frac{1}{2}\right) \tan ^{-1} \frac{x}{2}+\frac{8}{7} \cdot\left(\frac{1}{5}\right) \tan ^{-1}\left(\frac{x}{5}\right)+C \\ &I=\frac{-1}{14} \tan ^{-1} \frac{x}{2}+\frac{8}{35} \tan ^{-1} \frac{x}{5}+C \end{aligned}