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

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

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

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

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

Solution steps

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

Multiply fractions:

a \cdot \frac{b}{c} = \frac{a \cdot b}{c}

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

Apply u-substitution

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

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

Apply Integration By Parts

: - \frac{e ^{u} u}{1 + u} - \int - e^{u} d u

= - \frac{e ^{u} u}{1 + u} - \int - e^{u} d u

\int - e^{u} d u = - e^{u}

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

Substitute back

u = - x

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

Simplify

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

Add a constant to the solution

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

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