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

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

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integral of (3e^x)/(e^{2x)+6e^x+9}

\int \frac{3 e ^{x}}{e ^{2 x} + 6 e ^{x} + 9} d x

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

Solution steps

\int \frac{3 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

= 3 \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

= 3 \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}}

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

Apply u-substitution

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

= 3 \cdot \int \frac{1}{v ^{2}} d v

Apply the Power Rule

: - \frac{1}{v}

= 3 (- \frac{1}{v})

Substitute back

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

Simplify

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

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

Add a constant to the solution

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

y^{^{'}} + 2 y = 16

\int x (x^{2} - 1)^{5} d x

\frac{d ^{2}}{d x ^{2}} (e^{x} - x^{5})

\int 1 + (\frac{x ^{2}}{4} - \frac{1}{x ^{2}})^{2} d x

y^{^{''}} - 4 y^{^{'}} = e^{2 x}

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