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

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

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

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

e^{x} - 5 + 5 ln ∣ e^{x} - 5 ∣ + C

Solution steps

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

Apply u-substitution

: \int \frac{u + 5}{u} d u

= \int \frac{u + 5}{u} d u

Expand

\frac{u + 5}{u} : 1 + \frac{5}{u}

= \int 1 + \frac{5}{u} d u

Apply the Sum Rule:

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

= \int 1 d u + \int \frac{5}{u} d u

\int 1 d u = u

\int \frac{5}{u} d u = 5 ln ∣ u ∣

= u + 5 ln ∣ u ∣

Substitute back

u = e^{x} - 5

= e^{x} - 5 + 5 ln ∣ e^{x} - 5 ∣

Add a constant to the solution

= e^{x} - 5 + 5 ln ∣ e^{x} - 5 ∣ + C

f (x) = (7 x^{2} - 12 x) e^{x}

3 x^{2}, x^{2} + 6

\frac{d y}{d x} = 6 x y

2 e^{x}

\int_{0}^{1} 2 + 2 + x^{2} d x

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