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Evaluate the following integral:

Question:

Evaluate the following integral:

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

Solution:

Let, $I=\int \frac{x^{2}}{(x-1)\left(x^{2}+1\right)} d x$

Let $\frac{\mathrm{x}^{2}}{(\mathrm{x}-1)\left(\mathrm{x}^{2}+1\right)}=\frac{\mathrm{A}}{\mathrm{x}-1}+\frac{\mathrm{B}}{\mathrm{x}^{2}+1}$

$\Rightarrow x^{2}=A\left(x^{2}+1\right)+B(x-1)$

For, $\mathrm{x}=1, \mathrm{~A}=\frac{1}{2}$

For, $\mathrm{x}=0, \mathrm{~B}=\frac{1}{2}$

$\therefore \mathrm{I}=\frac{1}{2} \int \frac{\mathrm{dx}}{\mathrm{x}-1}+\frac{1}{2} \int \frac{\mathrm{dx}}{\mathrm{x}^{2}+1}$

$\Rightarrow \mathrm{I}=\frac{1}{2} \log |\mathrm{x}-1|+\frac{1}{2} \tan ^{-1} \mathrm{x}+\mathrm{c}$

Hence, $\int \frac{\mathrm{x}^{2}}{(\mathrm{x}-1)\left(\mathrm{x}^{2}+1\right)} \mathrm{dx}=\frac{1}{2} \log |\mathrm{x}-1|+\frac{1}{2} \tan ^{-1} \mathrm{x}+\mathrm{c}$

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