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

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

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

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

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

Solution steps

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

Apply u-substitution

: \int \frac{1}{( 1 - 3 u ) ^{2}} d u

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

Apply u-substitution

: \int - \frac{1}{3 v ^{2}} d v

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

Take the constant out:

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

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

Apply the Power Rule

: - \frac{1}{v}

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

Substitute back

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

Simplify

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

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

Add a constant to the solution

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

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