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$$\int_0^1 \frac{\arctan x}{1+x^{2}} dx = ?$$ I tried to evaluate it, but I do not know if it is good: $$\int_0^1\frac{\arctan x}{x^2+1}\,\mathrm{d}x=\left[\arctan x=t\Rightarrow \frac{\mathrm{d}x}{1+x^2}=\mathrm{d}t\right]=\int_0^{\frac{\pi}{4}}t \, \mathrm{d}t=\left[\frac{t^2}{2}\right]_0^{\frac{\pi}{4}}=\frac{\pi^2}{32}$$

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seems correct!! – Santosh Linkha Jan 21 '13 at 18:06

Here is another way to see the same calculation.

Notice that $\frac{\arctan x}{1+x^2}$ is of the form $u(x)u'(x)$ where $u(x) = \arctan(x)$. Hence, we have $$ \int_0^1\frac{\arctan x}{1+x^2}dx = \frac{1}{2}\left[(\arctan(x))^2\right]_0^1 = \frac{\pi^2}{32}. $$

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In general:

$$\int f'(x)f(x)^n\,dx=\frac{f(x)^{n+1}}{n+1}+C$$

So in our case, with

$$f(x)=\arctan x\,\,,\,\,f'(x)=\frac{1}{1+x^2}\Longrightarrow\int\frac{1}{1+x^2}\arctan x\,dx=\frac{1}{2}\arctan^2 x+C$$

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Your answer is correct. By the way, one solves equations and evaluates integrals.

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@Fre4kone In the same vein, please see the corrected title. – Did Jan 21 '13 at 18:22

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