Browse Questions

# Evaluate $\Large \int $$\large \frac{\log x }{(1+\log x)^2}$$dx$

Substitute $u = \log x$ and $du = \large\frac{1}{x}$$dx \Rightarrow I = \Large \int$$ \large\frac{ue^u}{(1+u^2)}$$du Integrating by parts, we get I = \Large\int$$e^u\;du - \large\frac{ ue^u}{1+u}$
Solving, we get $I = e^u - \large\frac{ue^u}{1+u}$ ($\large\int $$e^u = e^u) Substituting back u = \log x \rightarrow I = \large\frac{x}{1+\log x}$$ + c$
edited Mar 26, 2014 by balaji