I want to show that the definite integral [(ln x)/(1+x^2)] = 0. The definite integral's lower limit is 0 and the upper limit is ∞(infinity). The question asked the substitution x = 1/u to be used.
Thanks.
Let $\displaystyle I = \int_0^{+\infty} \frac{\ln x}{1 + x^2} \, dx$.
Hopefully you can make the substitution correctly to get $\displaystyle I = \int^0_{+\infty} \frac{- \ln u}{1 + \frac{1}{u^2}} \, \left( - \frac{du}{u^2}\right) = - \int_0^{+\infty} \frac{\ln u}{1 + u^2} \, du = -I$.
Therefore $\displaystyle 2I = 0 \Rightarrow I = 0$.