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Question Number 34562 by math khazana by abdo last updated on 08/May/18

$$findthevalueof{\int }_0^1\frac{arctanx}{{(1+x^2)}^2}dx$$

Commented by math khazana by abdo last updated on 08/May/18

$$wehaveprovedthat$$$$\int \frac{arctanx}{{(1+x^2)}^2}dx=\frac{1}{4}{\left(arctanx\right)}^2+\frac{1}{4}sin\left(2arctanx\right)+k\Rightarrow$$$${\int }_0^1\frac{arctanx}{{(1+x^2)}^2}dx={[\frac{1}{4}{\left(arctanx\right)}^2+\frac{1}{4}sin\left(2arctanx\right)]}_0^1$$$$=\frac{1}{4}{\left(\frac{\pi }{4}\right)}^2+\frac{1}{4}sin\left(\frac{\pi }{2}\right)=\frac{{\pi }^2}{64}+\frac{1}{4}.$$

Answered by tanmay.chaudhury50@gmail.com last updated on 08/May/18

$$t={tan}^{-1}x,\frac{dt}{dx}=\frac{1}{1+x^2}$$$$=\underset{0}{\stackrel{\mathrm{\Pi }/4}{\int }}\frac{t}{{(1+{tan}^{2}t)}^2}.(1+{tan}^{2}t)dt$$$$=\underset{0}{\stackrel{\mathrm{\Pi }/4}{\int }}t.{cos}^{2}tdt$$$$=\underset{0}{\stackrel{\mathrm{\Pi }/4}{\int }}t(1+cos2t)/2dt$$$$=\frac{1}{2}\underset{0}{\stackrel{\mathrm{\Pi }/4}{\int }}tdt+\frac{1}{2}\underset{0}{\stackrel{\mathrm{\Pi }/4}{\int }}tcos2tdt$$$$=\frac{1}{2}\mid \frac{t^2}{2}{\mid }_0^{\mathrm{\Pi }/4}+I_2$$$$=\frac{1}{4}.\frac{\prod ^2}{16}=\frac{\prod ^2}{64}$$$$I_2=\frac{1}{2}{\int }_0^{\mathrm{\Pi }/4}tcos2tdt$$$$\frac{1}{2}\mid \frac{tsin2t}{2}{\mid }_0^{\mathrm{\Pi }/4}-I_3$$$$\int tcos2tdt$$$$=t\frac{sin2t}{2}-\int \{\frac{dt}{dt}.\int cos2tdt\}dt$$$$=t\frac{sin2t}{2}-\int \frac{sin2t}{2}dt$$$$=t\frac{sin2t}{2}+\frac{cos2t}{4}$$$$so{I}_3=\mid \frac{tsin2t}{2}+\frac{cos2t}{4}{\mid }_0^{\mathrm{\Pi }/4}$$

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