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What is $\int \frac{x^{4}+1}{x^{2}+1} d x$ equal to?
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The correct answer is:
$\frac{x^{3}}{3}-x+2 \tan ^{-1} x+c$
$\int \frac{x^{4}+1}{x^{2}+1} d x=\int \frac{x^{4}+2-1}{x^{2}+1} d x=\int\left(\frac{x^{4}-1}{x^{2}+1}+\frac{2}{x^{2}+1}\right) d x$
$=\int\left[\frac{\left(x^{2}-1\right)\left(x^{2}+1\right)}{x^{2}+1}+\frac{2}{x^{2}+1}\right] d x$
$=\int\left(x^{2}-1+\frac{2}{x^{2}+1}\right) d x=\frac{x^{3}}{3}-x+2 \tan ^{-1} x+c$
$=\int\left[\frac{\left(x^{2}-1\right)\left(x^{2}+1\right)}{x^{2}+1}+\frac{2}{x^{2}+1}\right] d x$
$=\int\left(x^{2}-1+\frac{2}{x^{2}+1}\right) d x=\frac{x^{3}}{3}-x+2 \tan ^{-1} x+c$
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