Evaluating indefinite integral Evaluate the following indefinite integral.
$$\int { \frac { x }{ 4+{ x }^{ 4 } }  }\,dx$$
In my homework hints, it says let $ u = x^2 $. But still i can't continue.
 A: Hint: You've substituted $u = x^2$ and found that your original integral becomes
$$
\int\frac{\sqrt u}{4+u^2} \frac{du}{2x},
$$
but you haven't completed the substitution; there's still an $x$ in your integrand. How can you rewrite the $2x$ below the $du$ as a function of $u$? Once you rewrite $2x$ in terms of $u$, you should be able to algebraically simplify further.
Hint 2: You now have it in terms of $u$. Good! Do you see any way to simplify the integral? It may help to rewrite it as
$$
\int\frac{\sqrt u}{2\sqrt{u}(4+u^2)}du.
$$
A: Hint: If $u=x^2$ then $x=\sqrt u$ and $du=2x\,dx$.
A: Solve:
\begin{eqnarray}
\int\frac{x}{4+x^{4}}dx&=&\frac{1}{2}\int\frac{du}{4+u^2}; \text{ if $u=x^{2}$}\\
&=& \frac{1}{2}(\frac{1}{2}\arctan{\frac{u}{2}})\\
&=&\frac{1}{4}\arctan{\frac{x^{2}}{2}+C}
\end{eqnarray}
A: $\newcommand{\+}{^{\dagger}}%
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\begin{align}
\color{#0000ff}{\large\int{x\,\dd x \over 4 + x^{4}}}&=
\int x\pars{{1 \over x^{2} - 2\ic} - {1 \over x^{2} + 2\ic}}\,{1 \over 4\ic}\,\dd x
=
\half\,\Im\int{x\,\dd x \over x^{2} - 2\ic}
\\[3mm]&={1 \over 4}\,\Im\ln\pars{x^{2} - 2\ic}=
{1 \over 4}\,\arctan\pars{-2 \over \phantom{-}x^{2}}= \color{#0000ff}{\large -\,{1 \over 4}\,\arctan\pars{2 \over x^{2}}}
+ \mbox{"a constant"}
\\[3mm]&= \color{#0000ff}{\large {1 \over 4}\,\arctan\pars{x^{2} \over 2}}
+ \mbox{"some constant"}
\end{align}

Let's check it:
\begin{align}
\totald{}{x}\bracks{-\,{1 \over 4}\,\arctan\pars{2 \over x^{2}}}
&=
-\,{1 \over 4}\,
{1 \over \pars{2/x^{2}}^{2} + 1}\,\bracks{2\,\pars{-\,{2 \over x^{3}}}}
=
-\,{1 \over 4}\,
{x^{4} \over 4 + x^{4}}\,\pars{-4 \over \phantom{-}x^{3}}
\\[3mm]&={x \over 4 + x^{4}}
\end{align}

