How do you integrate #int ln(1+x^2)# by parts from #[0,1]#?

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Answer 1 · 2017-02-14T00:29:26

# int_0^1 ln(1+x^2) dx =ln2 -2 +pi/2#

We can take #x# as finite part:

#int_0^1 ln(1+x^2) dx = [xln(1+x^2)]_0^1 - int_0^1 x d(ln(1+x^2))#

#(1) int_0^1 ln(1+x^2) dx =ln2 - 2int_0^1 (x^2dx)/(1+x^2)#

Now solving the resulting integral:

#int_0^1 (x^2dx)/(1+x^2) = int_0^1 (x^2+1-1)/(1+x^2)dx#

#int_0^1 (x^2dx)/(1+x^2) = int_0^1 (1-1/(1+x^2))dx#

#int_0^1 (x^2dx)/(1+x^2) = int_0^1 dx - int_0^1 dx/(1+x^2)#

#int_0^1 (x^2dx)/(1+x^2) = [x]_0^1-[arctanx]_0^1#

#int_0^1 (x^2dx)/(1+x^2) = 1-pi/4#

Substitute in (1):

# int_0^1 ln(1+x^2) dx =ln2 -2 +pi/2#