Question

question in attachment

Answer 1

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\[\int\frac{2+\ln e^x}{e^{2x}}e^x\,dx=\int\frac{2+x}{e^x}\,dx.\]