How do you compute $$\int_{0}^1 \frac{\arctan x }{1+x} dx$$
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The answer is pi times ln2 over 8, no clue how they got that... – imranfat Mar 18 '13 at 19:42
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have you tried integration by part? Then you need to do $\int_0^1 \dfrac{\ln(1+x)}{1+x^2}dx$ – Yimin Mar 18 '13 at 19:52
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Using integration by parts $$\int_0^1 \frac{\arctan x}{1+x} dx = \arctan(x) \ln(1+x)|_0^1 - \int_0^1 \frac{\ln (1+x)}{1+x^2}dx$$
The former part is $\displaystyle \frac{\pi}{ 4} \ln 2 $ and the latter part is $\displaystyle \frac{\pi}{8} \ln 2$ which is answered here.
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Denominator of the given integral is linear (typo?), but your explanation makes sense, that link did it !! ThanX – imranfat Mar 18 '13 at 20:03
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$$ \begin{align} \int_0^1\frac{\arctan(x)}{1+x}\,\mathrm{d}x &=\int_0^{\pi/4}\frac{\theta}{1+\tan(\theta)}\,\sec^2(\theta)\,\mathrm{d}\theta\tag{1}\\[6pt] &=\int_0^{\pi/4}\frac{\theta\,\mathrm{d}\theta}{\cos(\theta)\,(\cos(\theta)+\sin(\theta))}\tag{2}\\[6pt] &=\int_0^{\pi/4}\frac{(\frac\pi4-\theta)\,\mathrm{d}\theta}{\cos(\theta)\,(\cos(\theta)+\sin(\theta))}\tag{3}\\[6pt] &=\frac\pi8\int_0^{\pi/4}\frac{\mathrm{d}\theta}{\cos(\theta)\,(\cos(\theta)+\sin(\theta))}\tag{4}\\[6pt] &=\frac\pi8\int_0^{\pi/4}\frac{\sec^2(\theta)}{1+\tan(\theta)}\,\mathrm{d}\theta\tag{5}\\[6pt] &=\frac\pi8\int_0^1\frac1{1+x}\mathrm{d}x\tag{6}\\[6pt] &=\frac\pi8\Big[\log(1+x)\Big]_0^1\tag{7}\\[12pt] &=\frac\pi8\log(2)\tag{8} \end{align} $$ $(1):$ $x=\tan(\theta)$
$(3):$ $\theta\mapsto\frac\pi4-\theta$
$(4):$ since $(2)=(3)$ we have $(3)=\frac{(2)+(3)}{2}$
$(6):$ $x=\tan(\theta)$
robjohn
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