Sol042806

Find exact value:

$\int_0^1 \frac{\log(1+x)\,dx}{1+x^2}\qquad(1)$

Solution I by landen Sometimes it is helpful to parameterize a definite integral so you get a function to work with. You differentiate and then integrate back. You just have to know to try this:

$\mbox{Define: }f(a)=\int_{0}^{1}{{{\log \left(a\,x+1\right)}\over{x^2+1}}\;dx};\qquad f(0)=0\qquad(2)$

${{d}\over{d\,a}}\,f\left(a\right)=\int_{0}^{1}{{{x}\over{\left(a\,x +1\right)\,\left(x^2+1\right)}}\;dx}\qquad(3)$

${{d}\over{d\,a}}\,f\left(a\right)=-{{\log \left(a+1\right)}\over{a^ 2+1}}+{{\pi\,a}\over{4\,a^2+4}}+{{2\,\log 2}\over{4\,a^2+4}}\qquad(4)$

$\int_0^1{{d}\over{d\,a}}\,f\left(a\right)\;da=-\int_0^1{{\log \left(a+1\right)}\over{a^ 2+1}}\;da+\int_0^1{{\pi\,a}\over{4\,a^2+4}}+{{2\,\log 2}\over{4\,a^2+4}}\;da\qquad(5)$

$f(1)=-f(1)+{{\pi\,\log 2}\over{4}}\qquad(6)$

$f(1)={{\pi\,\log 2}\over{8}}\qquad(7)$

$\square$

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