What is the integral of (e^{3x})/((1-2e^{3x))^2} ?

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

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integral of (e^{3x})/((1-2e^{3x))^2}

\int \frac{e ^{3 x}}{( 1 - 2 e ^{3 x} ) ^{2}} d x

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

Solution steps

\int \frac{e ^{3 x}}{( 1 - 2 e ^{3 x} ) ^{2}} d x

Apply u-substitution

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

= \int - \frac{1}{6 u ^{2}} d u

Take the constant out:

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

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

Apply the Power Rule

: - \frac{1}{u}

= - \frac{1}{6} (- \frac{1}{u})

Substitute back

u = 1 - 2 e^{3 x}

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

Simplify

- \frac{1}{6} (- \frac{1}{1 - 2 e ^{3 x}}) : \frac{1}{6 ( 1 - 2 e ^{3 x} )}

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

Add a constant to the solution

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

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

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

f (x) = x^{4} - 2 x^{2} + 1, at x = 2

\int_{0}^{π} sin (x) sin (3 x) d x

\frac{x ^{2}}{x ^{2} + 3}

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