Question

integral

Answer 1

\[e ^{2x}/\sqrt{1-e ^{4x}}\]

Answer 2

the inverse sinx is\[1/\sqrt{1-x ^{2}}\]

Answer 3

substitute u = e^2x
du = 2e^2x dx

Answer 4

what do i take out to make the \[e ^{4x}\] look like \[e ^{2x}\]

Answer 5

e^4x = (e^2x)^2
so if u = e^2x, then e^4x = u^2

Answer 6

oh ok thanks