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$$\int_0^\infty\dfrac{1}{(x+1)(x^2+1)}\text dx $$

I need to integrate the above. I had tried to decompose given integrand into two fractions, but I realized this is not always available.

Any hint or advice to handle it?

lioness99a
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Daschin
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  • Set $x=\tan y$ and use https://math.stackexchange.com/questions/439851/evaluate-the-integral-int-frac-pi2-0-frac-sin3x-sin3x-cos3xdx/439856#439856 – lab bhattacharjee Jul 06 '17 at 07:51
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    You can do a partial fractions decomposition and see that $$\frac{1}{(x+1)(x^2+1)} = -\frac{x}{2(x^2+1)} + \frac{1}{2(x^2+1)} + \frac{1}{2(x+1)}$$ and then you can integrate term by term – Osama Ghani Jul 06 '17 at 07:53
  • @Daschin you might have to do some logarithm simplification in the antiderivative to see that the improper integral converges, so watch out for that :) – Osama Ghani Jul 06 '17 at 07:58

3 Answers3

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Let $x=1/t$ to obtain $$I=\int_0^{\infty}\frac{tdt}{(t+1)(t^2+1)}=\int_0^{\infty}\frac{(t+1-1)dt}{(t+1)(t^2+1)}=-I+\int_0^{\infty}\frac{dt}{t^2+1}$$ Hence $$I=\frac12\int_0^{\infty}\frac{dt}{t^2+1}=\frac12 \times \frac{\pi}2.$$

Math-fun
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$$\frac{a}{x+1}+\frac{b x+c}{x^2+1}$$ hence $$\frac{a x^2+a+b x^2+b x+c x+c}{(x+1) \left(x^2+1\right)}=\frac{1}{(x+1) \left(x^2+1\right)}$$ $$a+c=1,b+c=0,a+b=0\rightarrow a= \frac{1}{2},b= -\frac{1}{2},c= \frac{1}{2}$$ The integrand becomes $$\frac{1}{2} \left(\frac{1}{x^2+1}+\frac{1}{x+1}\right)-\frac{2 x}{4 \left(x^2+1\right)}$$ which gives $$\left[\frac{1}{4} \log \left(\frac{(x+1)^2}{x^2+1}\right)+\frac{1}{2} \arctan(x)\right]_0^\infty=\frac{\pi}{4}$$

Raffaele
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As commented let $x=\tan(t) $ implying $dx=\sec^2 (t)dt $ so we have $I=\int _0^{\frac {\pi}{2}} \frac {dt}{1+\tan (t)} $ . Now its very well known that $I=\int _0^{\frac {\pi}{2}}\frac {dt}{1+\tan^a (t)}=\frac {\pi}{4} $ where $a\in R $

Archis Welankar
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