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The value of $\int \frac{x^{2}+1}{x^{2}-1} d x$ is
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Verified Answer
The correct answer is:
$x+\log \left(\frac{x-1}{x+1}\right)+c$
Let $I=\int \frac{x^{2}+1}{x^{2}-1} d x=\int \frac{x^{2}+1-1+1}{x^{2}-1} d x$
$$
\begin{aligned}
&\Rightarrow I=\int \frac{x^{2}-1}{x^{2}-1} d x+\int \frac{2}{x^{2}-1} d x \\
&\Rightarrow I=\int 1 d x+2 \int \frac{1}{x^{2}-1} d x \\
&\Rightarrow I=x+2 \cdot \frac{1}{2} \log \left(\frac{x-1}{x+1}\right)+c \\
&\Rightarrow I=x+\log \left(\frac{x-1}{x+1}\right)+c
\end{aligned}
$$
$$
\begin{aligned}
&\Rightarrow I=\int \frac{x^{2}-1}{x^{2}-1} d x+\int \frac{2}{x^{2}-1} d x \\
&\Rightarrow I=\int 1 d x+2 \int \frac{1}{x^{2}-1} d x \\
&\Rightarrow I=x+2 \cdot \frac{1}{2} \log \left(\frac{x-1}{x+1}\right)+c \\
&\Rightarrow I=x+\log \left(\frac{x-1}{x+1}\right)+c
\end{aligned}
$$
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