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How do I solve this? $$\int_0^\infty\frac{\ln(1+{x^2})}{1+{x^2}}$$ I know the answer is $\pi\ln2$.

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closed as off-topic by Carl Mummert, Watson, tatan, TravisJ, Qwerty Sep 2 '16 at 19:25

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    $\begingroup$ Posts which merely state a problem are discouraged on this site. Please improve your post by adding additional information. Where did the integral arise? Why is it of interest? What have you attempted already? $\endgroup$ – Carl Mummert Sep 2 '16 at 11:41
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    $\begingroup$ this integral was often solved on this forum $\endgroup$ – Dr. Sonnhard Graubner Sep 2 '16 at 11:44
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    $\begingroup$ There is search button on the top $\endgroup$ – Piotr Benedysiuk Sep 2 '16 at 12:00
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Meh.

$$\int_0^{+\infty} \frac{\ln(1+x^2)}{1 + x^2}\ \text{d}x = \frac{1}{2}\int_{-\infty}^{+\infty} \frac{\ln(1+x^2)}{1 + x^2}\ \text{d}x$$

Now set

$$x=\tan y$$

and you get:

$$-\frac{4}{2} \int^{\pi/2}_0 \ln(\cos y)\ \text{d}y = \pi \ln 2$$

Other References

Evaluating $\int_0^{\infty}\frac{\ln(x^2+1)}{x^2+1}dx$

But I prefer my method.

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