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Integrate the function
$\frac{3 x}{1+2 x^4}$
$\frac{3 x}{1+2 x^4}$
Solution:
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Verified Answer
Put $x^2=t$, so that $2 x d x=d t \Rightarrow x d x=\frac{d t}{2}$
$\begin{aligned}
&\therefore \quad \int \frac{3 x}{1+2 x^4} d x=\frac{1}{2} \int \frac{d t}{1+2 t^2}=\frac{3}{4} \int \frac{d t}{\left(\frac{1}{\sqrt{2}}\right)^2+t^2} \\
&=\frac{3}{2 \sqrt{2}} \tan ^{-1}(\sqrt{2 t)}+\mathrm{C} \\
&=\frac{3}{2 \sqrt{2}} \tan ^{-1}\left(\sqrt{2 x^2}\right)+\mathrm{C}
\end{aligned}$
$\begin{aligned}
&\therefore \quad \int \frac{3 x}{1+2 x^4} d x=\frac{1}{2} \int \frac{d t}{1+2 t^2}=\frac{3}{4} \int \frac{d t}{\left(\frac{1}{\sqrt{2}}\right)^2+t^2} \\
&=\frac{3}{2 \sqrt{2}} \tan ^{-1}(\sqrt{2 t)}+\mathrm{C} \\
&=\frac{3}{2 \sqrt{2}} \tan ^{-1}\left(\sqrt{2 x^2}\right)+\mathrm{C}
\end{aligned}$
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