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#### explain solution RD Sharma class 12 Chapter 18 Indefinite Integrals exercise 18.30 question 64

$\frac{19}{2 \sqrt{2}} \tan ^{-1} \frac{x}{\sqrt{2}}-\frac{39}{\sqrt{3}} \tan ^{-1} \frac{x}{\sqrt{3}}+\frac{67}{4} \tan ^{-1} \frac{x}{2}+C$

Hint:

To solve this integration, we use partial fraction method

Given:

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

Explanation:

Let

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

Put $x^{2}= y$

$I=\int \frac{4 y^{2}+3}{(y+2)(y+3)(y+4)}$

Let

\begin{aligned} &\frac{4 y^{2}+3}{(y+2)(y+3)(y+4)}=\frac{A}{y+2}+\frac{B}{y+3}+\frac{C}{y+4} \\ &4 y^{2}+3=A(y+3)(y+4)+B(y+2)(y+4)+C(y+2)(y+3) \end{aligned}

\begin{aligned} &y=-3 \\ &39=-B \\ &y=-4 \\ &67=C(-2)(-1) \\ &C=\frac{67}{2} \\ &y=-2 \\ &19=2 A \\ &A=\frac{19}{2} \end{aligned}

\begin{aligned} &I=\frac{19}{2} \int \frac{d x}{x^{2}+2}-39 \int \frac{d x}{x^{2}+3}+\frac{67}{2} \int \frac{d x}{x^{2}+4} \\ &I=\frac{19}{2 \sqrt{2}} \tan ^{-1} \frac{x}{\sqrt{2}}-\frac{39}{\sqrt{3}} \tan ^{-1} \frac{x}{\sqrt{3}}+\frac{67}{4} \tan ^{-1} \frac{x}{2}+C \end{aligned}