What is the integral of (xe^x)/((x+1)^2) ?

The integral of (xe^x)/((x+1)^2) is -(e^xx)/(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 Integration By Parts

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

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

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

\frac{d}{d x} ((5 x^{2} - 8 x) e^{x})

x^{^{''}} - 0 x^{^{'}} + x = 0

\int 6 d x

\frac{\partial}{\partial y} (\frac{- y}{x ^{2} + y ^{2}})

\int_{- \infty}^{0} e^{- x} d x

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