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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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    "Solve" is the wrong word. It should say "how to find" or "how to evaluate".2012-02-08

4 Answers 4

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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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    I got it !! Thank you so much !!2012-02-08
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$\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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    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 me2012-02-08