What is the integral of (e^{3t})/((e^{3t)+2)^{-4}} ?

The integral of (e^{3t})/((e^{3t)+2)^{-4}} is 1/15 (e^{3t}+2)^5+C

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integral of (e^{3t})/((e^{3t)+2)^{-4}}

\int \frac{e ^{3 t}}{( e ^{3 t} + 2 ) ^{- 4}} d t

\frac{1}{15} (e^{3 t} + 2)^{5} + C

Solution steps

\int \frac{e ^{3 t}}{( e ^{3 t} + 2 ) ^{- 4}} d t

Simplify

\frac{e ^{3 t}}{( e ^{3 t} + 2 ) ^{- 4}} : e^{3 t} (e^{3 t} + 2)^{4}

= \int e^{3 t} (e^{3 t} + 2)^{4} d t

Apply u-substitution

: \int \frac{u ^{4}}{3} d u

= \int \frac{u ^{4}}{3} d u

Take the constant out:

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

= \frac{1}{3} \cdot \int u^{4} d u

Apply the Power Rule

: \frac{u ^{5}}{5}

= \frac{1}{3} \cdot \frac{u ^{5}}{5}

Substitute back

u = e^{3 t} + 2

= \frac{1}{3} \cdot \frac{( e ^{3 t} + 2 ) ^{5}}{5}

Simplify

\frac{1}{3} \cdot \frac{( e ^{3 t} + 2 ) ^{5}}{5} : \frac{1}{15} (e^{3 t} + 2)^{5}

= \frac{1}{15} (e^{3 t} + 2)^{5}

Add a constant to the solution

= \frac{1}{15} (e^{3 t} + 2)^{5} + C

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What is the integral of (e^{3t})/((e^{3t)+2)^{-4}} ?
The integral of (e^{3t})/((e^{3t)+2)^{-4}} is 1/15 (e^{3t}+2)^5+C