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$\int \frac{x^{2}-1}{x^{4}+3 x^{2}+1} d x(x>0)$ is
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Verified Answer
The correct answer is:
$\tan ^{-1}\left(x+\frac{1}{x}\right)+C$
Let $I=\int \frac{x^{2}-1}{x^{4}+3 x^{2}+1} d x$
$=\int \frac{1-1 / x^{2}}{x^{2}+3+1 / x^{2}} d x$
$=\int \frac{1-1 / x^{2}}{\left(x^{2}+\frac{1}{x^{2}}\right)+3} d x$
$=\int \frac{1-1 / x^{2}}{\left(x+\frac{1}{x}\right)^{2}-2+3} d x$
$=\int \frac{1-1 / x^{2}}{\left(x+\frac{1}{x}\right)^{2}+1} d x$
Let $x+\frac{1}{x}=t$
$\Rightarrow \quad\left(1-\frac{1}{x^{2}}\right) d x=d t$
$\Rightarrow \quad I=\int \frac{d t}{t^{2}+1}$
$=\tan ^{-1} t+C$
$=\tan ^{-1}\left(x+\frac{1}{x}\right)+C\left[\because t=x+\frac{1}{x}\right]$
$=\int \frac{1-1 / x^{2}}{x^{2}+3+1 / x^{2}} d x$
$=\int \frac{1-1 / x^{2}}{\left(x^{2}+\frac{1}{x^{2}}\right)+3} d x$
$=\int \frac{1-1 / x^{2}}{\left(x+\frac{1}{x}\right)^{2}-2+3} d x$
$=\int \frac{1-1 / x^{2}}{\left(x+\frac{1}{x}\right)^{2}+1} d x$
Let $x+\frac{1}{x}=t$
$\Rightarrow \quad\left(1-\frac{1}{x^{2}}\right) d x=d t$
$\Rightarrow \quad I=\int \frac{d t}{t^{2}+1}$
$=\tan ^{-1} t+C$
$=\tan ^{-1}\left(x+\frac{1}{x}\right)+C\left[\because t=x+\frac{1}{x}\right]$
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