What is the integral of (4e^x)/(e^{2x)+8e^x+16} ?

The integral of (4e^x)/(e^{2x)+8e^x+16} is -4/(e^x+4)+C

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integral of (4e^x)/(e^{2x)+8e^x+16}

\int \frac{4 e ^{x}}{e ^{2 x} + 8 e ^{x} + 16} d x

- \frac{4}{e ^{x} + 4} + C

Solution steps

\int \frac{4 e ^{x}}{e ^{2 x} + 8 e ^{x} + 16} d x

Take the constant out:

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

= 4 \cdot \int \frac{e ^{x}}{e ^{2 x} + 8 e ^{x} + 16} d x

Apply u-substitution

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

= 4 \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}}

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

Apply u-substitution

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

= 4 \cdot \int \frac{1}{v ^{2}} d v

Apply the Power Rule

: - \frac{1}{v}

= 4 (- \frac{1}{v})

Substitute back

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

Simplify

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

= - \frac{4}{e ^{x} + 4}

Add a constant to the solution

= - \frac{4}{e ^{x} + 4} + C

\frac{\partial}{\partial x} (- x e^{\frac{x}{y}})

x \to 2 lim (13 - 2 x)

x \to 0 lim (8 x + 4 h - 3)

\frac{d}{d x} (1 - \frac{x ^{2}}{4})

\frac{\partial}{\partial v} (v sin (u))

What is the integral of (4e^x)/(e^{2x)+8e^x+16} ?
The integral of (4e^x)/(e^{2x)+8e^x+16} is -4/(e^x+4)+C