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

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

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

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

- \frac{1}{3 ( e ^{3 x} + 4 )} + C

Solution steps

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

Expand

e^{3 x} (e^{3 x} + 4)^{- 2} : \frac{e ^{3 x}}{e ^{6 x} + 8 e ^{3 x} + 16}

= \int \frac{e ^{3 x}}{e ^{6 x} + 8 e ^{3 x} + 16} d x

Apply u-substitution

: \int \frac{1}{3 ( u ^{2} + 8 u + 16 )} d u

= \int \frac{1}{3 ( u ^{2} + 8 u + 16 )} 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 \frac{1}{u ^{2} + 8 u + 16} d u

Take the partial fraction of

\frac{1}{u ^{2} + 8 u + 16} : \frac{1}{( u + 4 ) ^{2}}

= \frac{1}{3} \cdot \int \frac{1}{( u + 4 ) ^{2}} d u

Apply u-substitution

: \int \frac{1}{v ^{2}} d v

= \frac{1}{3} \cdot \int \frac{1}{v ^{2}} d v

Apply the Power Rule

: - \frac{1}{v}

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

Substitute back

= \frac{1}{3} (- \frac{1}{e ^{3 x} + 4})

Simplify

\frac{1}{3} (- \frac{1}{e ^{3 x} + 4}) : - \frac{1}{3 ( e ^{3 x} + 4 )}

= - \frac{1}{3 ( e ^{3 x} + 4 )}

Add a constant to the solution

= - \frac{1}{3 ( e ^{3 x} + 4 )} + C

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