What is the integral of (e^{3x})/(e^x-5) ?

The integral of (e^{3x})/(e^x-5) is ((e^x-5)^2)/2+10(e^x-5)+25ln|e^x-5|+C

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integral of (e^{3x})/(e^x-5)

\int \frac{e ^{3 x}}{e ^{x} - 5} d x

\frac{( e ^{x} - 5 ) ^{2}}{2} + 10 (e^{x} - 5) + 25 ln ∣ e^{x} - 5 ∣ + C

Solution steps

\int \frac{e ^{3 x}}{e ^{x} - 5} d x

Apply u-substitution

: \int \frac{u ^{2}}{u - 5} d u

= \int \frac{u ^{2}}{u - 5} d u

Apply u-substitution

: \int \frac{( v + 5 ) ^{2}}{v} d v

= \int \frac{( v + 5 ) ^{2}}{v} d v

Expand

\frac{( v + 5 ) ^{2}}{v} : v + 10 + \frac{25}{v}

= \int v + 10 + \frac{25}{v} d v

Apply the Sum Rule:

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

= \int v d v + \int 10 d v + \int \frac{25}{v} d v

\int v d v = \frac{v ^{2}}{2}

\int 10 d v = 10 v

\int \frac{25}{v} d v = 25 ln ∣ v ∣

= \frac{v ^{2}}{2} + 10 v + 25 ln ∣ v ∣

Substitute back

= \frac{( e ^{x} - 5 ) ^{2}}{2} + 10 (e^{x} - 5) + 25 ln ∣ e^{x} - 5 ∣

Add a constant to the solution

= \frac{( e ^{x} - 5 ) ^{2}}{2} + 10 (e^{x} - 5) + 25 ln ∣ e^{x} - 5 ∣ + C

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\int \frac{x}{cos ^{2} ( 4 x ^{2} + \frac{1}{a} )} d x

What is the integral of (e^{3x})/(e^x-5) ?
The integral of (e^{3x})/(e^x-5) is ((e^x-5)^2)/2+10(e^x-5)+25ln|e^x-5|+C