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

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

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integral of (2xe^x)/((x+1)^2)

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

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

Solution steps

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

Take the constant out:

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

= 2 \cdot \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

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

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

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

Simplify

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

Add a constant to the solution

= 2 (- \frac{e ^{x} x}{x + 1} + e^{x}) + C

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

sin (x^{2}), 0

y^{^{''}} + y^{^{'}} - 2 y = sin (x)

\frac{d y}{d x} - y = e^{2 x} y^{3}

\frac{d ^{2} y}{d x ^{2}} + y = x^{2} + 2

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