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

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

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integral of (2e^x)/(e^{2x)+10e^x+25}

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

- \frac{2}{e ^{x} + 5} + C

Solution steps

\int \frac{2 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

= 2 \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

= 2 \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}}

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

Apply u-substitution

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

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

Apply the Power Rule

: - \frac{1}{v}

= 2 (- \frac{1}{v})

Substitute back

= 2 (- \frac{1}{e ^{x} + 5})

Simplify

2 (- \frac{1}{e ^{x} + 5}) : - \frac{2}{e ^{x} + 5}

= - \frac{2}{e ^{x} + 5}

Add a constant to the solution

= - \frac{2}{e ^{x} + 5} + C

\int \frac{x ^{3}}{( x - 1 ) ^{4}} d x

\int_{2}^{\infty} y e^{- 9 y} d y

\frac{d ^{2}}{d x ^{2}} ((x^{3} + 2 x)^{6})

n = 0 \sum \infty (\frac{x}{n !})^{n}

y = (x + 4)^{5} (x + 1)^{2}

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