What is the integral of (e^x)/(e^{2x)+8e^x+15} ?

The integral of (e^x)/(e^{2x)+8e^x+15} is 1/2 ln|e^x+3|-1/2 ln|e^x+5|+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of (e^x)/(e^{2x)+8e^x+15}

\int \frac{e ^{x}}{e ^{2 x} + 8 e ^{x} + 15} d x

\frac{1}{2} ln ∣ e^{x} + 3 ∣ - \frac{1}{2} ln ∣ e^{x} + 5 ∣ + C

Solution steps

\int \frac{e ^{x}}{e ^{2 x} + 8 e ^{x} + 15} d x

Apply u-substitution

: \int \frac{1}{u ^{2} + 8 u + 15} d u

= \int \frac{1}{u ^{2} + 8 u + 15} d u

Take the partial fraction of

\frac{1}{u ^{2} + 8 u + 15} : \frac{1}{2 ( u + 3 )} - \frac{1}{2 ( u + 5 )}

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

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

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

\int \frac{1}{2 ( u + 3 )} d u = \frac{1}{2} ln ∣ u + 3 ∣

\int \frac{1}{2 ( u + 5 )} d u = \frac{1}{2} ln ∣ u + 5 ∣

= \frac{1}{2} ln ∣ u + 3 ∣ - \frac{1}{2} ln ∣ u + 5 ∣

Substitute back

u = e^{x}

= \frac{1}{2} ln ∣ e^{x} + 3 ∣ - \frac{1}{2} ln ∣ e^{x} + 5 ∣

Add a constant to the solution

= \frac{1}{2} ln ∣ e^{x} + 3 ∣ - \frac{1}{2} ln ∣ e^{x} + 5 ∣ + C

\int \frac{x}{( 5 - 4 x - x ^{2} ) ^{3}} d x

\int e^{x} x d x

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

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

\int 1 + 2 x^{2} d x

What is the integral of (e^x)/(e^{2x)+8e^x+15} ?
The integral of (e^x)/(e^{2x)+8e^x+15} is 1/2 ln|e^x+3|-1/2 ln|e^x+5|+C