# 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.

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

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

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

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

Hint: If $u=x^2$ then $x=\sqrt u$ and $du=2x\,dx$.

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

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}


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}