#### explain solution RD Sharma class 12 Chapter 18 Indefinite Integrals exercise 18.30 question 69

$\frac{1}{4 \sqrt{3}} \tan ^{-1} \frac{x}{\sqrt{3}}+\frac{27}{40} \log \frac{x-5}{x+5}+C$

Hint:

To solve this integration, we use partial fraction method

Given:

$\int \frac{\left(x^{2}+1\right)\left(x^{2}+4\right)}{\left(x^{2}+3\right)\left(x^{2}-5\right)} d x \\$

Explanation:

Let

\begin{aligned} &I=\int \frac{\left(x^{2}+1\right)\left(x^{2}+4\right)}{\left(x^{2}+3\right)\left(x^{2}-5\right)} d x \\ &I=\int \frac{x^{4}+5 x^{2}+4}{x^{4}-2 x^{2}-15} d x \\ &I=\int\left(1+\frac{7 x^{2}+19}{x^{4}-2 x^{2}-15}\right) d x \\ &I=x-\int \frac{7 x^{2}+19}{x^{4}-2 x^{2}-15} d x \end{aligned}

$I=x-I_{1}$                           (1)

Where

\begin{aligned} &I_{1}=\int \frac{7 x^{2}+19}{x^{4}-2 x^{2}-15} d x \\ &x^{2}=y \\ &\frac{7 y+19}{y^{2}-2 y-15}=\frac{7 y+19}{(y+3)(y-5)}=\frac{A}{y+3}+\frac{B}{y-5} \\ &7 y+19=A(y-5)+B(y+3) \end{aligned}

\begin{aligned} &\text { At } y=5 \\ &54=8 B \\ &B=\frac{27}{4} \end{aligned}

\begin{aligned} &\text { At } y=-3 \\ &-2=-8 A \\ &A=\frac{1}{4} \end{aligned}

\begin{aligned} &\int \frac{7 x^{2}+19}{x^{4}-2 x^{2}-15}=\frac{1}{4} \int \frac{1}{x^{2}+3} d x+\frac{27}{4} \int \frac{1}{x^{2}-5} d x \\ &=\frac{1}{4} \cdot \frac{1}{\sqrt{3}} \tan ^{-1} \frac{x}{\sqrt{3}}+\frac{27}{4} \cdot \frac{1}{2 \times 5} \log \frac{x-5}{x+5}+C \\ &=\frac{1}{4 \sqrt{3}} \tan ^{-1} \frac{x}{\sqrt{3}}+\frac{27}{40} \log \frac{x-5}{x+5}+C \end{aligned}