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$$\int_{0}^{1} \frac{\log{(1+x)}}{x^{2}+1} \ dx $$

I tried substituting x with 1/t but couldn't find the answer. Can someone provide any hint?

As many suggested I substituted x with tan t but again I got stuck at $$\int_{0}^{\frac{π}{4}} \log{(1+tant)} \ dt $$

I finally solved it and got the correct answer.

  • 1
    $x=\tan \theta$ – 2017-01-21

2 Answers 2

-1

Hint -

Put $\tan^{-1} x = t$

Then x = tan t.

And $\frac{1}{1 + x^2} dx = dt$

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    All downvotes are welcome. But please tell the reason so that I can improve my mistake. – 2017-01-21
-1

Put $\displaystyle x = \frac{1-t}{1+t}$