You're asking this in the wrong section, but here's your procedure anyway: Consider$\int\frac{2+\ln u}{u^2}\,du.$Let $$u=e^x$$. Then, $$du=e^x\,dx$$. Substituting this into the integral yields$\int\frac{2+\ln e^x}{e^{2x}}e^x\,dx=\int\frac{2+x}{e^x}\,dx.$