What is the integral of (e^{13x})/((e^{13x)+5)^4} ?

The integral of (e^{13x})/((e^{13x)+5)^4} is -1/(39(e^{13x)+5)^3}+C

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integral of (e^{13x})/((e^{13x)+5)^4}

\int \frac{e ^{13 x}}{( e ^{13 x} + 5 ) ^{4}} d x

- \frac{1}{39 ( e ^{13 x} + 5 ) ^{3}} + C

Solution steps

\int \frac{e ^{13 x}}{( e ^{13 x} + 5 ) ^{4}} d x

Apply u-substitution

: \int \frac{e ^{u}}{13 ( e ^{u} + 5 ) ^{4}} d u

= \int \frac{e ^{u}}{13 ( e ^{u} + 5 ) ^{4}} d u

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= \frac{1}{13} \cdot \int \frac{e ^{u}}{( e ^{u} + 5 ) ^{4}} d u

Apply u-substitution

: \int \frac{1}{v ^{4}} d v

= \frac{1}{13} \cdot \int \frac{1}{v ^{4}} d v

Apply the Power Rule

: - \frac{1}{3 v ^{3}}

= \frac{1}{13} (- \frac{1}{3 v ^{3}})

Substitute back

= \frac{1}{13} (- \frac{1}{3 ( e ^{13 x} + 5 ) ^{3}})

Simplify

\frac{1}{13} (- \frac{1}{3 ( e ^{13 x} + 5 ) ^{3}}) : - \frac{1}{39 ( e ^{13 x} + 5 ) ^{3}}

= - \frac{1}{39 ( e ^{13 x} + 5 ) ^{3}}

Add a constant to the solution

= - \frac{1}{39 ( e ^{13 x} + 5 ) ^{3}} + C

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What is the integral of (e^{13x})/((e^{13x)+5)^4} ?
The integral of (e^{13x})/((e^{13x)+5)^4} is -1/(39(e^{13x)+5)^3}+C