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

The integral of (e^x)/(e^{2x)+6e^x+9} is -1/(e^x+3)+C

Popular >

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

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

- \frac{1}{e ^{x} + 3} + C

Solution steps

\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

= \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}}

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

Apply u-substitution

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

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

Apply the Power Rule

: - \frac{1}{v}

= - \frac{1}{v}

Substitute back

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

Add a constant to the solution

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

\frac{d}{d x} (e^{- x} \cdot sin (x))

\frac{d}{d x} (e^{\frac{1}{x ^{3}}})

\frac{d y}{d x} y = (y + 1) (2 x + 1), y (- 1) = 1

\int z^{2} (z^{3} - 1)^{6} d z

\int p (p + 6)^{5} d p

What is the integral of (e^x)/(e^{2x)+6e^x+9} ?
The integral of (e^x)/(e^{2x)+6e^x+9} is -1/(e^x+3)+C