What is the integral of (2e^x)/(e^{2x)+6e^x+9} ?

The integral of (2e^x)/(e^{2x)+6e^x+9} is -2/(e^x+3)+C

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integral of (2e^x)/(e^{2x)+6e^x+9}

\int \frac{2 e ^{x}}{e ^{2 x} + 6 e ^{x} + 9} d x

- \frac{2}{e ^{x} + 3} + C

Solution steps

\int \frac{2 e ^{x}}{e ^{2 x} + 6 e ^{x} + 9} 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} + 6 e ^{x} + 9} d x

Apply u-substitution

: \int \frac{1}{u ^{2} + 6 u + 9} d u

= 2 \cdot \int \frac{1}{u ^{2} + 6 u + 9} d u

Take the partial fraction of

\frac{1}{u ^{2} + 6 u + 9} : \frac{1}{( u + 3 ) ^{2}}

= 2 \cdot \int \frac{1}{( u + 3 ) ^{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} + 3})

Simplify

2 (- \frac{1}{e ^{x} + 3}) : - \frac{2}{e ^{x} + 3}

= - \frac{2}{e ^{x} + 3}

Add a constant to the solution

= - \frac{2}{e ^{x} + 3} + C

f (x) = 4 x^{2} cos (x) - 9 x

\frac{d}{d x} (700 x^{2} - 0.2 x)

\int \frac{5}{6} sec (x) tan (x) d x

y^{2} d x + 2 x y d y = 0

What is the integral of (2e^x)/(e^{2x)+6e^x+9} ?
The integral of (2e^x)/(e^{2x)+6e^x+9} is -2/(e^x+3)+C