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How can i calculate the Integral $\displaystyle \int_{0}^{1}\frac{\tan^{-1}(x)}{1+x}dx$

$\bf{My\; Trial::}$ Let $\displaystyle I(\alpha) = \int_{0}^{\alpha}\frac{\tan^{-1}(\alpha x)}{1+x}dx$

Now $\displaystyle \frac{d}{d\alpha}I(\alpha) = \frac{d}{d\alpha}\int_{0}^{\alpha}\frac{\tan^{-1}(\alpha x)}{1+x}dx = \int_{0}^{\alpha}\frac{d}{d\alpha}\frac{\tan^{-1}(\alpha x)}{1+x}dx = \int_{0}^{\alpha}\frac{x}{(1+\alpha^2 x^2)(1+x)}dx$

Now How can I solve after that

Help me

Thanks

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  • $\begingroup$ You have a mistake in the bounds. You should gave the bounds on the last line be 0 to 1, not $\alpha$. $\endgroup$ – abnry Jun 8 '14 at 18:53
  • $\begingroup$ It's simpler to just integrate from $0$ to $1$. Otherwise you have a missing term in the derivative. $\endgroup$ – Robert Israel Jun 8 '14 at 19:01
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    $\begingroup$ Try integration by parts: $I=\bigg[\ln(1+x)\arctan x\bigg]_0^1-\displaystyle\int_0^1\frac{\ln(1+x)}{1+x^2}dx = \frac\pi4\cdot\ln2-\frac\pi8\cdot\ln2=$ $=\dfrac\pi8\cdot\ln2$. $\endgroup$ – Lucian Jun 8 '14 at 19:05
  • $\begingroup$ But when you do get $\dfrac{d}{d\alpha} I(\alpha)$, you'll still have an integration to do to recover $I(\alpha)$. $\endgroup$ – Robert Israel Jun 8 '14 at 19:07
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Let us denote the considered integral by $I$. The change of variables $x=\frac{1-t}{1+t}$ shows that $$ I=\int_0^1\frac{1}{1+t}\tan^{-1}\left(\frac{1-t}{1+t}\right)dt $$ But if $f(t)=\tan^{-1}\left(\frac{1-t}{1+t}\right)+\tan^{-1}t$, then it is easy to see that $f'(t)=0$ so $f(t)=f(0)=\pi/4$ for $0\leq t\leq 1$, hence $$ I=\int_0^1\frac{\frac{\pi}{4}-\tan^{-1}t}{1+t}dt=\frac{\pi}{4}\ln(2)-I $$ So, $I=\dfrac{\pi}{8}\ln2$.

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  • $\begingroup$ Beautiful solution! $\endgroup$ – user111187 Jun 8 '14 at 19:35
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Put$x=\tan \theta$. Then $$I=\int_{0}^1\frac{\tan^{-1}x}{1+x}dx=\int_{0}^{\pi/4}\frac{\theta\sec^2\theta}{1+\tan \theta}d\theta\\=\int_{0}^{\pi/4}\frac{\theta\sec\theta}{\sin\theta+\cos\theta}d\theta=\int_{0}^{\pi/4}\frac{2\theta}{\sin 2\theta+\cos2\theta+1}d\theta\\=\frac{1}{2}\int_{0}^{\pi/2}\frac{\theta}{\sin\theta+\cos\theta+1}d\theta=\pi/8\int_{0}^{\pi/2}\frac{d\theta}{\sin\theta+\cos\theta+1}=\pi/8J$$ where $$J=\int_{0}^{\pi/2}\frac{d\theta}{\sin\theta+\cos\theta+1}$$ Put $t=\tan \theta/2$ to get $$J=\int_{0}^{1}\frac{dz}{(1+z)}=\ln 2$$ So, $I=\frac{\pi}{8}\ln 2$

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Hint: Apply inegration by part to get;

$$= (\dfrac{1}{x^2+1})(\dfrac{1}{x+1})-\int \dfrac{1}{x^2+1}(\dfrac{-1}{(x+1)^2}) dx$$

By partial fraction, $$\dfrac{1}{x^2+1}(\dfrac{-1}{(x+1)^2})=\dfrac{x}{2(x^2+1)}-\dfrac{1}{2(x+1)^2}-\dfrac{1}{2(x+1)}$$

From that you can easily integrate this.

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    $\begingroup$ This just makes things worse, you should use partial fractions now. $\endgroup$ – Rene Schipperus Jun 8 '14 at 18:51
  • $\begingroup$ @ReneSchipperus: Unless there is a way of calculating the definite integral without finding indefinite integral, it is not a bad way. $\endgroup$ – mesel Jun 8 '14 at 19:01
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You can use the following partial fractions to continue. Let $$ \frac{x}{(1+\alpha^2x^2)(1+x)}=\frac{Ax+B}{1+\alpha^2x^2}+\frac{C}{1+x}. $$ It is easy to get $$ A=B=\frac{1}{2-\alpha^2}, C=-A. $$ I think you can do the rest.

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