Question

evaluate the following integral.
int\frac{e^{4x}}{e^{4x}-5e^{-4x}}dx
int\frac{e^{4x}}{e^{4x}-5e^{-4x}}dx=square
(use parentheses to clearly denote the argument of each function.)

Explanation:

Step1: Use substitution

Let \(u = e^{4x}-5e^{- 4x}\), then \(du=(4e^{4x}+20e^{-4x})dx = 4(e^{4x} + 5e^{-4x})dx\). And the integrand \(\frac{e^{4x}}{e^{4x}-5e^{-4x}}dx=\frac{1}{4}\frac{4e^{4x}}{e^{4x}-5e^{-4x}}dx\).

Step2: Rewrite integral in terms of u

\(\int\frac{e^{4x}}{e^{4x}-5e^{-4x}}dx=\frac{1}{4}\int\frac{du}{u}\)

Step3: Integrate with respect to u

We know that \(\int\frac{du}{u}=\ln|u|+C\), so \(\frac{1}{4}\int\frac{du}{u}=\frac{1}{4}\ln|u|+C\)

Step4: Substitute back u

Substituting \(u = e^{4x}-5e^{-4x}\) back, we get \(\frac{1}{4}\ln|e^{4x}-5e^{-4x}|+C\)

Answer:

\(\frac{1}{4}\ln|e^{4x}-5e^{-4x}|+C\)