What is the integral of (5e^x)/(e^{2x)+6e^x+9} ?

The integral of (5e^x)/(e^{2x)+6e^x+9} is -5/(e^x+3)+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of (5e^x)/(e^{2x)+6e^x+9}

\int \frac{5 e ^{x}}{e ^{2 x} + 6 e ^{x} + 9} d x

- \frac{5}{e ^{x} + 3} + C

Solution steps

\int \frac{5 e ^{x}}{e ^{2 x} + 6 e ^{x} + 9} d x

Take the constant out:

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

= 5 \cdot \int \frac{e ^{x}}{e ^{2 x} + 6 e ^{x} + 9} d x

Apply u-substitution

: \int \frac{1}{u ^{2} + 6 u + 9} d u

= 5 \cdot \int \frac{1}{u ^{2} + 6 u + 9} d u

Take the partial fraction of

\frac{1}{u ^{2} + 6 u + 9} : \frac{1}{( u + 3 ) ^{2}}

= 5 \cdot \int \frac{1}{( u + 3 ) ^{2}} d u

Apply u-substitution

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

= 5 \cdot \int \frac{1}{v ^{2}} d v

Apply the Power Rule

: - \frac{1}{v}

= 5 (- \frac{1}{v})

Substitute back

= 5 (- \frac{1}{e ^{x} + 3})

Simplify

5 (- \frac{1}{e ^{x} + 3}) : - \frac{5}{e ^{x} + 3}

= - \frac{5}{e ^{x} + 3}

Add a constant to the solution

= - \frac{5}{e ^{x} + 3} + C

\int_{0}^{2} (4 x - x^{3}) d x

\frac{( 4 s - 3 )}{( s ^{2} + 8 s + 24 )}

x + 2 y = 7, x + 8 = y^{2}

\int 14 x^{6} y^{- 4} - 12 x^{5} y^{- 3} d x

x \to 3 lim (12 - x - 3)

What is the integral of (5e^x)/(e^{2x)+6e^x+9} ?
The integral of (5e^x)/(e^{2x)+6e^x+9} is -5/(e^x+3)+C