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

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

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

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

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

Solution steps

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

Apply u-substitution

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

Factor

- u - 1 : - (u + 1)

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

(- (u + 1))^{2} = (u + 1)^{2}

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

Apply Integration By Parts

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

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

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

Substitute back

u = - x

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

Simplify

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

Add a constant to the solution

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

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