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Evaluate the following integral:

Question:

Evaluate the following integral:

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

Solution:

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

Let $x^{2}=y$

$\therefore \frac{4 \mathrm{x}^{4}+3}{\left(\mathrm{x}^{2}+2\right)\left(\mathrm{x}^{2}+3\right)\left(\mathrm{x}^{2}+4\right)}=\frac{4 \mathrm{y}^{2}+3}{(\mathrm{y}+2)(\mathrm{y}+3)(\mathrm{y}+4)}$

Let, $\frac{4 y^{2}+3}{(y+2)(y+3)(y+4)}=\frac{A}{y+2}+\frac{B}{y+3}+\frac{C}{y+4}$

$\Rightarrow 4 y^{2}+3=A(y+3)(y+4)+B(y+2)(y+4)+C(y+2)(y+3)$

For $y=-2, A=\frac{19}{2}$

For $y=-3, B=-39$

For $y=-4, C=\frac{67}{2}$

Thus, $I=\frac{19}{2} \int \frac{d x}{x^{2}+2}-39 \int \frac{d x}{x^{2}+3}+\frac{67}{2} \int \frac{d x}{x^{2}+4}$

$\Rightarrow I=\frac{19}{2 \sqrt{2}} \tan ^{-1}\left(\frac{x}{\sqrt{2}}\right)-\frac{39}{\sqrt{3}} \tan ^{-1}\left(\frac{x}{\sqrt{3}}\right)+\frac{67}{4} \tan ^{-1}\left(\frac{x}{2}\right)+c$

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