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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.

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    $\begingroup$ Why can't you continue? Can you show us how far you get before getting stuck? What happens when you perform the substitution $u = x^2$? $\endgroup$ – Dan Jan 13 '14 at 0:42
  • $\begingroup$ $ \int { \frac { x }{ 4+{ ({ x }^{ 2 }) }^{ 2 } } } \quad dx\\ \\ \int { \frac { \sqrt { u } }{ 4+{ u }^{ 2 } } } \quad dx\\ \\ \int { \frac { \sqrt { u } }{ 4+{ u }^{ 2 } } } \quad \frac { du }{ 2x } $ Then what ? $\endgroup$ – Out Of Bounds Jan 13 '14 at 0:52
  • $\begingroup$ So far so good! I'll reply with an answer containing more suggestions. $\endgroup$ – Dan Jan 13 '14 at 0:54
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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

$$ \int\frac{\sqrt u}{2\sqrt{u}(4+u^2)}du. $$

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  • $\begingroup$ $ \int { \frac { \sqrt { u } }{ 4+{ u }^{ 2 } } } \quad \frac { du }{ 2\sqrt { u } } $ . Actually I'm a bit rusty in algebra so i still can't figure it out. $\endgroup$ – Out Of Bounds Jan 13 '14 at 1:00
  • $\begingroup$ @Tennisman: See my edit. $\endgroup$ – Dan Jan 13 '14 at 1:05
  • $\begingroup$ Yeah now i got it :) Thank you Dan :) $\endgroup$ – Out Of Bounds Jan 13 '14 at 1:08
  • $\begingroup$ Happy to help! :) $\endgroup$ – Dan Jan 13 '14 at 2:59
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Hint: If $u=x^2$ then $x=\sqrt u$ and $du=2x\,dx$.

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  • $\begingroup$ $ \int { \frac { x }{ 4+{ ({ x }^{ 2 }) }^{ 2 } } } \quad dx\\ \\ \int { \frac { \sqrt { u } }{ 4+{ u }^{ 2 } } } \quad dx\\ \\ \int { \frac { \sqrt { u } }{ 4+{ u }^{ 2 } } } \quad \frac { du }{ 2x } $ Then what ? $\endgroup$ – Out Of Bounds Jan 13 '14 at 0:53
  • $\begingroup$ Yes. $\displaystyle\int\frac{x}{4+u^2}\,\frac{du}{2x}$ $\endgroup$ – Berci Jan 13 '14 at 23:00
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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}

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$\newcommand{\+}{^{\dagger}}% \newcommand{\angles}[1]{\left\langle #1 \right\rangle}% \newcommand{\braces}[1]{\left\lbrace #1 \right\rbrace}% \newcommand{\bracks}[1]{\left\lbrack #1 \right\rbrack}% \newcommand{\ceil}[1]{\,\left\lceil #1 \right\rceil\,}% \newcommand{\dd}{{\rm d}}% \newcommand{\down}{\downarrow}% \newcommand{\ds}[1]{\displaystyle{#1}}% \newcommand{\equalby}[1]{{#1 \atop {= \atop \vphantom{\huge A}}}}% \newcommand{\expo}[1]{\,{\rm e}^{#1}\,}% \newcommand{\fermi}{\,{\rm f}}% \newcommand{\floor}[1]{\,\left\lfloor #1 \right\rfloor\,}% \newcommand{\half}{{1 \over 2}}% \newcommand{\ic}{{\rm i}}% \newcommand{\iff}{\Longleftrightarrow} \newcommand{\imp}{\Longrightarrow}% \newcommand{\isdiv}{\,\left.\right\vert\,}% \newcommand{\ket}[1]{\left\vert #1\right\rangle}% \newcommand{\ol}[1]{\overline{#1}}% \newcommand{\pars}[1]{\left( #1 \right)}% \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\pp}{{\cal P}}% \newcommand{\root}[2][]{\,\sqrt[#1]{\,#2\,}\,}% \newcommand{\sech}{\,{\rm sech}}% \newcommand{\sgn}{\,{\rm sgn}}% \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}} \newcommand{\ul}[1]{\underline{#1}}% \newcommand{\verts}[1]{\left\vert\, #1 \,\right\vert}$ \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}
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