The value of the integral _ 2^ 3e^2xe^4x-1,dx is equal to:

Question

The value of the integral $\displaystyle\int_{\ln 2}^{\ln 3}\frac{e^{2x}}{e^{4x}-1}\,dx$ is equal to:

Accepted Answer

Substitute $t=e^{2x}$, $dt = 2e^{2x}dx$; limits become $t=4$ to $t=9$:

$$I = \frac12\int_4^9\frac{dt}{t^2-1} = \frac14\left[\ln\left(\frac{t-1}{t+1}\right)\right]_4^9 = \frac14\left[\ln\frac45 - \ln\frac35\right] = \frac14\ln\left(\frac43\right)$$

Solutions · 1