How to integrate : e^(x) dx / (e^(2x) - 25)

Let u = e^x, so du = e^x dx

Then, we have du/(u^2 - 25) = du/(u + 5)(u - 5)

Now, 1/(u + 5)(u - 5) = A/(u + 5) + B/(u - 5)

so, 1 = A(u - 5) + B(u + 5)

when u = 5, B = 1/10 and when u = - 5, A = -1/10

Then integrate, 1/10(u - 5) - 1/10(u + 5) du

so, 1/10[ln (u - 5) - ln (u + 5)] + C

i.e. 1/10[ln (u - 5)/(u + 5)] + C

=> 1/10[ln (e^x - 5)/(e^x + 5)] + C

:)>

Notice that e^(2x) = [e^x]^2. The denominator is the difference of squares. So it can be factored as (e^x - 5)(e^x + 5).

That suggests using partial fractions, writing 1/(e^x - 5)(e^x + 5) as [A/(e^x - 5)] + [B/(e^x + 5)].

If it helps first substitute y = e^x. So you're rewriting y/(y^2 - 25) in terms of partial fractions.