What is the integral of (3e^x)/(e^{2x)+8e^x+16} ?

The integral of (3e^x)/(e^{2x)+8e^x+16} is -3/(e^x+4)+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of (3e^x)/(e^{2x)+8e^x+16}

\int \frac{3 e ^{x}}{e ^{2 x} + 8 e ^{x} + 16} d x

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

Solution steps

\int \frac{3 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

= 3 \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

= 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}}

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

Apply u-substitution

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

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

Apply the Power Rule

: - \frac{1}{v}

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

Substitute back

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

Simplify

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

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

Add a constant to the solution

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

\int 4 x - x^{2} d x

n = 0 \sum \infty n e^{n}

\int x + 10 d x

\int_{0}^{4} (4 x - x^{2}) d x

\frac{\partial}{\partial x} (\frac{x ^{2}}{y ^{3}})

What is the integral of (3e^x)/(e^{2x)+8e^x+16} ?
The integral of (3e^x)/(e^{2x)+8e^x+16} is -3/(e^x+4)+C