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I was solving $\int x^2 \arctan(x) \;dx $ I set $u=x^2$, $dv= \arctan(x)$, so I could get $du=2x$, $v=x\arctan(x)-\frac12\ln(1+x^{2})$.

From $\int x^2 \arctan(x)\;dx = uv - \int v \; du$

I got

$$\begin{align*}&\int x^2 \arctan(x) \; dx =\\ &x^3\arctan(x) -\frac12\ln(1+x^{2})-\int 2x\left[x^{2}\arctan x -\frac12\ln(1+x^2)\right]\;dx \end{align*}$$

and simplified if; then I got

$$3\int x^2\arctan(x) \;dx = x^3 \arctan(x)-\frac12\ln(1+x^2)+\int x\ln(1+x^2) \;dx$$

after that I use $w=\ln(1+x^2) \; dv =dx$ to find $\int x\ln(1+ x^2)$

but I got

$$x^2 \arctan(x)-\int 2x\arctan(x) \; dx$$

If I got $$x^2 \arctan(x)-\int 2x^2 \arctan(x)\;dx$$ instead, it would be easy to solve the question....

How can I solve this question and if you find any my mistake could you post this wall ??

Thank you !

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1  
It surprises me that you would try setting $dv=\arctan(x)$. Wouldn't it make more sense to set $u=\arctan(x)$? After all, $\frac{d}{dx}\arctan(x)$ is a rational function, as is $v$ if $dv=x^2\,dx$; so you will end up having to solve the integral $$\int\frac{x^3}{1+x^2}\,dx$$which can be solved using long division and an easy substitution. –  Arturo Magidin Feb 7 '12 at 22:29
    
In short: try $u=\arctan(x)$, $dv=x^2\,dx$ instead; that will get rid of that annoying $\arctan(x)$, and not keep it around nor replace it with $\ln$. The moral is: after integration by parts, the objective is for the new integral to be simpler, or at least no more complicated, than the original one. You traded a hard integral for a harder integral, and that's not the direction you want to go. –  Arturo Magidin Feb 7 '12 at 22:31
    
This is a case in which the LIATE rule works. –  Américo Tavares Feb 7 '12 at 22:57
1  
"Solve" is the wrong word. It should say "how to find" or "how to evaluate". –  Michael Hardy Feb 8 '12 at 19:30

4 Answers 4

up vote 4 down vote accepted

Let $u=\arctan x$, $dv = x^2\,dx$. Then $du = \frac{dx}{1+x^2}$, $v = \frac{1}{3}x^3$, so $$\begin{align*} \int x^2\arctan x\,dx &= \frac{1}{3}x^3\arctan x - \frac{1}{3}\int\frac{x^3}{1+x^2}\,dx\\ &= \frac{1}{3}x^3\arctan x - \frac{1}{3}\int\left(\frac{x^3+x}{1+x^2}-\frac{x}{1+x^2}\,dx\right)\\ &= \frac{1}{3}x^3\arctan x - \frac{1}{3}\int x\,dx + \frac{1}{3}\int\frac{x}{1+x^2}\,dx\\ &= \frac{1}{3}x^3\arctan x - \frac{1}{6}x^2 +\frac{1}{3}\int\frac{\frac{1}{2}\,du}{u} &\quad&(u=1+x^2)\\ &=\frac{1}{3}x^3\arctan x - \frac{1}{6}x^2 + \frac{1}{6}\int\frac{du}{u}\\ &= \frac{1}{3}x^3\arctan x - \frac{1}{6}x^2 + \frac{1}{6}\ln|u| + C\\ &= \frac{1}{3}x^3\arctan x - \frac{1}{6}x^2 + \frac{1}{6}\ln|1+x^2|+C\\ &= \frac{1}{3}x^3\arctan x - \frac{1}{6}x^2 + \frac{1}{6}\ln(1+x^2)+C. \end{align*}$$ If after integration by parts/substitution, the resulting integral is harder than the one you started with, then it's time to go back and try a different integration by parts/substitution.

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@Américo: Thanks for the correction. –  Arturo Magidin Feb 7 '12 at 22:55
    
I got it !! Thank you so much !! –  Ryu Feb 8 '12 at 0:57

$$\int{ x^2 \cdot \tan^{-1} x} dx = $$

$$ \tan^{-1} x dx = du $$

$$ x\cdot\tan^{-1} x - \frac{1}{2}\log(x^2+1) = u $$

$$ x^2 = v $$

$$ 2xdx = dv $$

$$\int{ x^2 \cdot \tan^{-1} x} dx = x^2\left(x\cdot\tan^{-1} x - \frac{1}{2}\log(x^2+1) \right)-\int 2x\left(x\cdot\tan^{-1} x - \frac{1}{2}\log(x^2+1) \right)dx$$

$$I = {x^2}\left( {x\cdot{{\tan }^{ - 1}}x - \frac{1}{2}\log ({x^2} + 1)} \right) - 2I + \int {x\log \left( {{x^2} + 1} \right)} dx$$

$$3I = {x^2}\left( {x\cdot{{\tan }^{ - 1}}x - \frac{1}{2}\log ({x^2} + 1)} \right) + \frac{1}{2}\int {\log u} du$$

$$3I = {x^2}\left( {x\cdot{{\tan }^{ - 1}}x - \frac{1}{2}\log ({x^2} + 1)} \right) + \frac{{{x^2} + 1}}{2}\left[ {\log \left( {{x^2} + 1} \right) - 1} \right]$$

$$I = \frac{{{x^2}}}{3}\left( {x\cdot{{\tan }^{ - 1}}x - \frac{1}{2}\log ({x^2} + 1)} \right) + \frac{{{x^2} + 1}}{6}\left[ {\log \left( {{x^2} + 1} \right) - 1} \right]$$

This simplifies to $$I = \frac{{{x^3}}}{3}\cdot{\tan ^{ - 1}}x + \frac{1}{6}\log \left( {{x^2} + 1} \right) - \frac{{{x^2}}}{6} + C $$

which coincides with WA's solution.

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Suggestion: Since $\frac{\mathrm{d}}{\mathrm{d}x}\arctan(x)=\frac{1}{1+x^2}$, it seems to be simpler to let $u=\arctan(x)$ and $\mathrm{d}v=x^2\,\mathrm{d}x$. Then you get $$ \begin{align} \frac{x^3}{3}\arctan(x)-\frac13\int\frac{x^3}{1+x^2}\mathrm{d}x &=\frac{x^3}{3}\arctan(x)-\frac13\int\left(x-\frac{x}{1+x^2}\right)\mathrm{d}x\\ &=\frac{x^3}{3}\arctan(x)-\frac16x^2+\frac16\int\frac{\mathrm{d}(1+x^2)}{1+x^2}\\ &=\frac{x^3}{3}\arctan(x)-\frac16x^2+\frac16\log(1+x^2)+C \end{align} $$

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We have $$ \int x^2 \arctan x\;dx. $$

Let $$ \begin{align} u & = \arctan x, \\ \\ du & = \frac{dx}{x^2+1}, \\ \\ dv & = x^2 \;dx, \\ \\ v & = \frac{x^3}{3}. \end{align} $$

Then $$ \begin{align} \int u\;dv & = uv - \int v\;du = \frac{x^3}{3}\arctan x - \int \frac{x^3\;dx}{3(x^2+1)} = \frac{x^3}{3}\arctan x - \frac 1 3\int \left(x- \frac{x}{x^2+1}\right)\;dx \\ \\ & = \frac{x^3}{3}\arctan x - \frac{x^2}{6} + \frac 16 \log(x^2+1) +C. \end{align} $$

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First Arturo, then me, then you. Each separated by a minute from the next. :-) –  robjohn Feb 8 '12 at 1:32
    
But not before me :D - 22:31. Why don't I get upvotes? =( –  Pedro Tamaroff Feb 8 '12 at 2:58
    
@Peter: Your answer used the harder integration by parts; I was grouping the three successive answers using the same integration by parts :-) –  robjohn Feb 8 '12 at 6:08
    
ha well, I just wrote what he asked for. –  Pedro Tamaroff Feb 8 '12 at 12:09
    
Im sorry about confusion. I just begin to use this site so I thought I could vote up for every one because everyone's answer was super good! Forgive me –  Ryu Feb 8 '12 at 20:41

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