What is the integral of (e^x)/(e^{2x)+4e^x-5} ?

The integral of (e^x)/(e^{2x)+4e^x-5} is 1/6 ln|e^x-1|-1/6 ln|e^x+5|+C

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integral of (e^x)/(e^{2x)+4e^x-5}

\int \frac{e ^{x}}{e ^{2 x} + 4 e ^{x} - 5} d x

\frac{1}{6} ln ∣ e^{x} - 1 ∣ - \frac{1}{6} ln ∣ e^{x} + 5 ∣ + C

Solution steps

\int \frac{e ^{x}}{e ^{2 x} + 4 e ^{x} - 5} d x

Take the partial fraction of

\frac{e ^{x}}{e ^{2 x} + 4 e ^{x} - 5} : \frac{1}{6 ( e ^{x} - 1 )} + \frac{5}{6 ( e ^{x} + 5 )}

= \int \frac{1}{6 ( e ^{x} - 1 )} + \frac{5}{6 ( e ^{x} + 5 )} d x

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= \int \frac{1}{6 ( e ^{x} - 1 )} d x + \int \frac{5}{6 ( e ^{x} + 5 )} d x

\int \frac{1}{6 ( e ^{x} - 1 )} d x = \frac{1}{6} (ln ∣ e^{x} - 1 ∣ - x)

\int \frac{5}{6 ( e ^{x} + 5 )} d x = \frac{1}{6} (x - ln ∣ e^{x} + 5 ∣)

= \frac{1}{6} (ln ∣ e^{x} - 1 ∣ - x) + \frac{1}{6} (x - ln ∣ e^{x} + 5 ∣)

Simplify

\frac{1}{6} (ln ∣ e^{x} - 1 ∣ - x) + \frac{1}{6} (x - ln ∣ e^{x} + 5 ∣) : \frac{1}{6} ln ∣ e^{x} - 1 ∣ - \frac{1}{6} ln ∣ e^{x} + 5 ∣

= \frac{1}{6} ln ∣ e^{x} - 1 ∣ - \frac{1}{6} ln ∣ e^{x} + 5 ∣

Add a constant to the solution

= \frac{1}{6} ln ∣ e^{x} - 1 ∣ - \frac{1}{6} ln ∣ e^{x} + 5 ∣ + C

What is the integral of (e^x)/(e^{2x)+4e^x-5} ?
The integral of (e^x)/(e^{2x)+4e^x-5} is 1/6 ln|e^x-1|-1/6 ln|e^x+5|+C