evaluate the definite intergral ∫0^1/2(8e2x)/(2e2x-1)dx

Question

evaluate the definite intergral ₀^1/2(8e^2x)/(2e^2x-1)dx

Answer 1

∫_{(from 0 to 1/2)}[ 8e^2x / (2e^2x - 1) ]dx

Let u = 2e^2x - 1 Then du = 4e^2xdx. So, 8e^2xdx = 2du

When x = 0, u = 1 and when x = 1/2, u = 2e-1.

So, the given integral is equivalent to ∫_{(from 1 to 2e-1)} (2/u)du = 2ln l u l_{(from 1 to 2e-1)}

2ln(2e-1) - 2ln1 = 2ln(2e-1)