Question

{log (1+x)}dx÷(1+x^2) integration

Answer 1

Here's a solution that uses simpler tools (or at least tools that I'm more familiar with):

I=∫1to0 log(1+x)1+x2. We change x into x=tan(t). Then t=arctanx, dt=11+x2dx, and the integral becomes:

I=∫π/4 to 0 log(1+tan(t))dt.

Now s=π4−t, ds=−dt, and the integral becomes:

I=−∫0 to π/4 log(1+tan(π4−s))ds=∫π40log(1+tan(π4−s))ds

Using tan(a+b)=tana−tanb1+tan(a)tan(b), we have

I=∫π/4 to 0 log(1+1−tans1+tans)ds=

∫π/4 to 0 log(21+tans)ds=∫π40(log(2)−log(1+tans))ds=∫π/4 to 0 log(2)ds−I=π/4 log(2)−I.

So I=π/4log(2)−I, hence I=π/8log(2).