integral 0 to 1 x(tan-1x)2dx

xtan-1xdx=x22tan-1x2-1+x2-1tan-1x1+x2dx=x22tan-1x2-tan-1x dx+tan-1x1+x2dx
tan-1x dx=xtan-1x-x1+x2dx=xtan-1x-12ln1+x2
x22tan-1x2+xtan-1x-12ln1+x2+tan-1x22
Evaluating this from 0 to 1 we get that:
π42+π4-12ln 2 

  • -14

pi/2-log2

  • 1

take 2 as second function and tan-1x as first functions and then solve by

By parts

  • -1
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