Question

Evaluate:∫(e^ax-e^-ax)/(e^ax+e^-ax)dx

Answer 1

solution

∫(e∧ax-e∧-ax)/(e∧ax+e∧-ax)

let u=e∧ax+e∧-ax)

du/dx=(aeax-ae-ax)

⇒dx=du/(ae∧ax-ae∧-ax)

then we substitute

∫(e∧ax-e∧-axu)/u*du/(ae∧ax-ae∧-ax)

⇒∫1/u*1/a

⇒=1/a∫1/u*du

=1/a(inu)+c

1/a(in(e∧ax+e∧-ax)+c