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If $\int \frac{x^2+1}{x^4+1} d x=f(x)+c$, then $f(x)$ is equal to
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Verified Answer
The correct answer is:
$\frac{1}{\sqrt{2}} \tan ^{-1}\left(\frac{x^2-1}{\sqrt{2} x}\right)$
$\int \frac{x^2+1}{x^4+1} d x=f(x)+c$
Let
$$
\begin{aligned}
& I=\int \frac{x^2+1}{x^4+1} d x=\int \frac{x^2\left(1+\frac{1}{x^2}\right)}{x^2\left(x^2+\frac{1}{x^2}\right)} d x \\
& I=\int \frac{\left(1+\frac{1}{x^2}\right)}{\left(x-\frac{1}{x}\right)^2+(\sqrt{2})^2} d x
\end{aligned}
$$
Let $x-\frac{1}{x}=t$
$\begin{aligned} & \Rightarrow\left(1+\frac{1}{x^2}\right) d x=d t \\ & \quad I=\int \frac{d t}{(\sqrt{2})^2+t^2}=\frac{1}{\sqrt{2}} \tan ^{-1}\left(\frac{t}{\sqrt{2}}\right)+c \\ & \quad I=\frac{1}{\sqrt{2}} \tan ^{-1}\left(\frac{x^2-1}{\sqrt{2} x}\right)+c=f(x)+c \\ & \therefore f(x)=\frac{1}{\sqrt{2}} \tan ^{-1}\left(\frac{x^2-1}{\sqrt{2} x}\right)\end{aligned}$
Let
$$
\begin{aligned}
& I=\int \frac{x^2+1}{x^4+1} d x=\int \frac{x^2\left(1+\frac{1}{x^2}\right)}{x^2\left(x^2+\frac{1}{x^2}\right)} d x \\
& I=\int \frac{\left(1+\frac{1}{x^2}\right)}{\left(x-\frac{1}{x}\right)^2+(\sqrt{2})^2} d x
\end{aligned}
$$
Let $x-\frac{1}{x}=t$
$\begin{aligned} & \Rightarrow\left(1+\frac{1}{x^2}\right) d x=d t \\ & \quad I=\int \frac{d t}{(\sqrt{2})^2+t^2}=\frac{1}{\sqrt{2}} \tan ^{-1}\left(\frac{t}{\sqrt{2}}\right)+c \\ & \quad I=\frac{1}{\sqrt{2}} \tan ^{-1}\left(\frac{x^2-1}{\sqrt{2} x}\right)+c=f(x)+c \\ & \therefore f(x)=\frac{1}{\sqrt{2}} \tan ^{-1}\left(\frac{x^2-1}{\sqrt{2} x}\right)\end{aligned}$
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