What is the integral of (e^{2t})/(1+6e^{2t)+9e^{4t}} ?

The integral of (e^{2t})/(1+6e^{2t)+9e^{4t}} is -1/(6(3e^{2t)+1)}+C

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integral of (e^{2t})/(1+6e^{2t)+9e^{4t}}

\int \frac{e ^{2 t}}{1 + 6 e ^{2 t} + 9 e ^{4 t}} d t

- \frac{1}{6 ( 3 e ^{2 t} + 1 )} + C

Solution steps

\int \frac{e ^{2 t}}{1 + 6 e ^{2 t} + 9 e ^{4 t}} d t

Apply u-substitution

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

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

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

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

Complete the square

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

= \frac{1}{2} \cdot \int \frac{1}{9 ( u + \frac{1}{3} ) ^{2}} d u

Apply u-substitution

: \int \frac{1}{9 v ^{2}} d v

= \frac{1}{2} \cdot \int \frac{1}{9 v ^{2}} d v

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= \frac{1}{2} \cdot \frac{1}{9} \cdot \int \frac{1}{v ^{2}} d v

Apply the Power Rule

: - \frac{1}{v}

= \frac{1}{2} \cdot \frac{1}{9} (- \frac{1}{v})

Substitute back

= \frac{1}{2} \cdot \frac{1}{9} (- \frac{1}{e ^{2 t} + \frac{1}{3}})

Simplify

\frac{1}{2} \cdot \frac{1}{9} (- \frac{1}{e ^{2 t} + \frac{1}{3}}) : - \frac{1}{6 ( 3 e ^{2 t} + 1 )}

= - \frac{1}{6 ( 3 e ^{2 t} + 1 )}

Add a constant to the solution

= - \frac{1}{6 ( 3 e ^{2 t} + 1 )} + C

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What is the integral of (e^{2t})/(1+6e^{2t)+9e^{4t}} ?
The integral of (e^{2t})/(1+6e^{2t)+9e^{4t}} is -1/(6(3e^{2t)+1)}+C