Question

Help me please, I have to solve an integral by substitution!

Answer 1

i'd just slog it out with \(u = e^x + 1 , \ e^x = u-1, \ du = e^x .dx \implies dx = \frac{du}{u-1}\)

leaving \(\int \frac{u-2}{u} \frac{1}{u-1} \ du\) and some partial fractions to fiddle about with

then remember to transform it back with \(u = e^x + 1\)