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

\int x \cdot sin (3 x) d x

x cos (π) x, x = 0

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

f (x) = \frac{1}{2 x + 3}

\int (x^{3} + 2 x)^{5} (6 x^{2} + 4) 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