#### explain solution RD Sharma class 12 Chapter 18 Indefinite Integrals exercise 18.30 question 63

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

Hint:

To solve this integration, we use partial fraction method

Given:

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

Explanation:

Let

\begin{aligned} &I=\int \frac{\left(x^{2}+1\right)\left(x^{2}+2\right)}{\left(x^{2}+3\right)\left(x^{2}+4\right)} d x \\ &I=\int \frac{x^{4}+3 x^{2}+2}{x^{4}+7 x^{2}+12} d x \\ &I=\int\left(1-\frac{\left(4 x^{2}+10\right)}{x^{4}+7 x^{2}+12}\right) d x \\ &I=\int 1 d x-\int \frac{4 x^{2}+10}{x^{4}+7 x^{2}+12} d x \\ &I=x-I_{1} \end{aligned}                (1)

Where

\begin{aligned} &I_{1}=\int \frac{4 x^{2}+10}{x^{4}+7 x^{2}+12} d x \\ &\frac{4 x^{2}+10}{x^{4}+7 x^{2}+12}=\frac{4 x^{2}+10}{\left(x^{2}+3\right)\left(x^{2}+4\right)} \end{aligned}              (i)

$4 x^{2}+10=(A x+B)\left(x^{2}+4\right)+(C x+D)\left(x^{2}+3\right)$

Comparing the coefficient

$x^{3}$ Coefficient

$0=A+C$                         (2)

$x^{2}$ Coefficient

$4=B+D$                         (3)

$x$ Coefficient

$0=4 A+3 C$                     (4)

Constant

$10=4 B+3 D$                    (5)

\begin{aligned} &\text { Term (2) } A=-C \text { put in (4) }\\ &0=-4 C+3 C\\ &C=0\\ &A=0 \end{aligned}

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

\begin{aligned} &\begin{array}{l} 4 B+4 D=16 \\ 4 B+3 D=10 \\ \hline D=6 \\ B+D=4 \\ B+6=4 \\ B=-2 \end{array} \end{aligned}

(i) Becomes

\begin{aligned} &I_{1}=\int \frac{-2}{x^{2}+3} d x+6 \int \frac{1}{x^{2}+4} d x \\ &I_{1}=-2 \cdot \frac{1}{\sqrt{3}} \tan ^{-1} \frac{x}{\sqrt{3}}+6 \cdot \frac{1}{2} \tan ^{-1} \frac{x}{2} \end{aligned}

\begin{aligned} &I_{1}=\frac{-2}{\sqrt{3}} \tan ^{-1} \frac{x}{\sqrt{3}}+3 \tan ^{-1} \frac{x}{2} \\ &I=x-I_{1} \\ &I=x+\frac{2}{\sqrt{3}} \tan ^{-1} \frac{x}{\sqrt{3}}-3 \tan ^{-1} \frac{x}{2}+C \end{aligned}