2x∫(0->1)f(x)dx+f(x)=ln(1+x^2)
∫(0->1) [2x∫(0->1)f(x)dx+f(x) ] dx =∫(0->1) ln(1+x^2) dx
∫(0->1)f(x)dx .[x^2](0->1) + ∫(0->1)f(x)dx =∫(0->1) ln(1+x^2) dx
∫(0->1)f(x)dx =(1/2)∫(0->1) ln(1+x^2) dx
= (1/2)[xln(1+x^2)](0->1) - ∫(0->1) x^2/(1+x^2) dx
= (1/2)ln2 - ∫(0->1)dx + ∫(0->1) dx/(1+x^2)
= (1/2)ln2 - 1 + [arctanx](0->1)
=(1/2)ln2 - 1 +π/4