Although the solution given in the video is by far the simplest, it may be of interest for some to see how Feynman's trick can apply to this question. Rewrite the original integral as integral_0^inf{dx (ln(1+x^3)-ln(1+x))/(1+x^2)/ln(x)} = F(3)-F(1), where F(a) = integral_0^inf{dx ln(1+x^a)/(1+x^2)/ln(x))}. Therefore, F'(a) = integral_0^inf{dx x^a/(1+x^2)/(1+x^a)} = integral_0^(pi/2){ds tan^a(s)/(1+tan^a(s))} = integral_0^(pi/2){ds sin^a(s)/(sin^a(s)+cos^a(s))} = integral_0^(pi/2){ds cos^a(s)/(sin^a(s)+cos^a(s))} = 1/2*integral_0^(pi/2){ds (sin^a(s)+cos^a(s))/(sin^a(s)+cos^a(s))} = pi/4. Thus, I = F(3)-F(1) = integral_1^3{da pi/4} = pi/2
Gorgeous!
This integral would very difficult without the inverse function substitution.
Although the solution given in the video is by far the simplest, it may be of interest for some to see how Feynman's trick can apply to this question.
Rewrite the original integral as integral_0^inf{dx (ln(1+x^3)-ln(1+x))/(1+x^2)/ln(x)} = F(3)-F(1), where F(a) = integral_0^inf{dx ln(1+x^a)/(1+x^2)/ln(x))}. Therefore, F'(a) = integral_0^inf{dx x^a/(1+x^2)/(1+x^a)} = integral_0^(pi/2){ds tan^a(s)/(1+tan^a(s))} = integral_0^(pi/2){ds sin^a(s)/(sin^a(s)+cos^a(s))} = integral_0^(pi/2){ds cos^a(s)/(sin^a(s)+cos^a(s))} = 1/2*integral_0^(pi/2){ds (sin^a(s)+cos^a(s))/(sin^a(s)+cos^a(s))} = pi/4. Thus, I = F(3)-F(1) = integral_1^3{da pi/4} = pi/2