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

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

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

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

- \frac{2}{e ^{x} + 4} + C

Solution steps

\int \frac{2 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

= 2 \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

= 2 \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}}

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

Simplify

2 (- \frac{1}{e ^{x} + 4}) : - \frac{2}{e ^{x} + 4}

= - \frac{2}{e ^{x} + 4}

Add a constant to the solution

= - \frac{2}{e ^{x} + 4} + C

x \to - 3 + lim (\frac{1}{x + 3})

f (x) = - x^{3} + 4 x^{2}, at x = 3

y = x^{3} - 2 x^{2} + 3, y = 3 x^{2} + 5 x - 22

y = x^{\frac{15}{14}}, y = 13 x^{\frac{1}{14}}

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