What is the integral of (e^{4v})/((1+3e^{4v))^2} ?

The integral of (e^{4v})/((1+3e^{4v))^2} is -1/(12(1+3e^{4v))}+C

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integral of (e^{4v})/((1+3e^{4v))^2}

\int \frac{e ^{4 v}}{( 1 + 3 e ^{4 v} ) ^{2}} d v

- \frac{1}{12 ( 1 + 3 e ^{4 v} )} + C

Solution steps

\int \frac{e ^{4 v}}{( 1 + 3 e ^{4 v} ) ^{2}} d v

Apply u-substitution

: \int \frac{1}{12 u ^{2}} d u

= \int \frac{1}{12 u ^{2}} d u

Take the constant out:

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

= \frac{1}{12} \cdot \int \frac{1}{u ^{2}} d u

Apply the Power Rule

: - \frac{1}{u}

= \frac{1}{12} (- \frac{1}{u})

Substitute back

u = 1 + 3 e^{4 v}

= \frac{1}{12} (- \frac{1}{1 + 3 e ^{4 v}})

Simplify

\frac{1}{12} (- \frac{1}{1 + 3 e ^{4 v}}) : - \frac{1}{12 ( 1 + 3 e ^{4 v} )}

= - \frac{1}{12 ( 1 + 3 e ^{4 v} )}

Add a constant to the solution

= - \frac{1}{12 ( 1 + 3 e ^{4 v} )} + C

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What is the integral of (e^{4v})/((1+3e^{4v))^2} ?
The integral of (e^{4v})/((1+3e^{4v))^2} is -1/(12(1+3e^{4v))}+C