What is the integral of (5e^x)/(e^{2x)+8e^x+16} ?

The integral of (5e^x)/(e^{2x)+8e^x+16} is -5/(e^x+4)+C

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integral of (5e^x)/(e^{2x)+8e^x+16}

\int \frac{5 e ^{x}}{e ^{2 x} + 8 e ^{x} + 16} d x

- \frac{5}{e ^{x} + 4} + C

Solution steps

\int \frac{5 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

= 5 \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

= 5 \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}}

= 5 \cdot \int \frac{1}{( u + 4 ) ^{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} + 4})

Simplify

5 (- \frac{1}{e ^{x} + 4}) : - \frac{5}{e ^{x} + 4}

= - \frac{5}{e ^{x} + 4}

Add a constant to the solution

= - \frac{5}{e ^{x} + 4} + C

\int (\frac{7}{x ^{4}}) d x

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What is the integral of (5e^x)/(e^{2x)+8e^x+16} ?
The integral of (5e^x)/(e^{2x)+8e^x+16} is -5/(e^x+4)+C