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

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

Popular >

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

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

- \frac{1}{2 ( e ^{2 x} + 4 )} + C

Solution steps

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

Apply u-substitution

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

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

Take the constant out:

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

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

Apply the Power Rule

: - \frac{1}{u}

= \frac{1}{2} (- \frac{1}{u})

Substitute back

u = e^{2 x} + 4

= \frac{1}{2} (- \frac{1}{e ^{2 x} + 4})

Simplify

\frac{1}{2} (- \frac{1}{e ^{2 x} + 4}) : - \frac{1}{2 ( e ^{2 x} + 4 )}

= - \frac{1}{2 ( e ^{2 x} + 4 )}

Add a constant to the solution

= - \frac{1}{2 ( e ^{2 x} + 4 )} + C

\int \frac{sec ^{2} ( x ) + sec ^{5} ( x )}{cot ( x )} d x

\int \frac{1}{x ^{2} + 8 - 6 x} d x

\int \frac{sec ^{3} ( \frac{1}{θ} ) tan ^{2} ( \frac{1}{θ} )}{θ ^{2}} d θ

\int \frac{1}{3 - 4 t ^{2}} d t

\int 2 x sinh (2 x) d x

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