What is the integral of (5e^x)/(e^{2x)+10e^x+25} ?

The integral of (5e^x)/(e^{2x)+10e^x+25} is -5/(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 (5e^x)/(e^{2x)+10e^x+25}

\int \frac{5 e ^{x}}{e ^{2 x} + 10 e ^{x} + 25} d x

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

Solution steps

\int \frac{5 e ^{x}}{e ^{2 x} + 10 e ^{x} + 25} 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} + 10 e ^{x} + 25} d x

Apply u-substitution

: \int \frac{1}{u ^{2} + 10 u + 25} d u

= 5 \cdot \int \frac{1}{u ^{2} + 10 u + 25} d u

Take the partial fraction of

\frac{1}{u ^{2} + 10 u + 25} : \frac{1}{( u + 5 ) ^{2}}

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

Simplify

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

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

Add a constant to the solution

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

x \to 2 lim (- (x))

y^{^{''}} = t

d y - (y - 4) 2 d x = 0

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

\frac{d}{d x} (ln (x + x^{2}))

What is the integral of (5e^x)/(e^{2x)+10e^x+25} ?
The integral of (5e^x)/(e^{2x)+10e^x+25} is -5/(e^x+5)+C