Question

evaluate integration of 1 to 0 log of 1 + X by x square DX​

Answer 1

Answer:

I need to evaluate this integral: ∫10log(x+1)1+x2dx.

I've tried t=log(x+1), t=x+1, but to no avail. I've noticed that:

∫10log(x+1)1+x2dx=∫10log(x+1)arctan′(x)dx=log(x+1)arctan(x)|x=1x=0−∫10arctan(x)x+1dx