What is the integral from 0 to 1 of (xe^x)/((1+x)^2) ?

The integral from 0 to 1 of (xe^x)/((1+x)^2) is e/2-1

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integral from 0 to 1 of (xe^x)/((1+x)^2)

\int_{0}^{1} \frac{x e ^{x}}{( 1 + x ) ^{2}} d x

\frac{e}{2} - 1

Decimal Notation

0.35914 \dots

Solution steps

\int_{0}^{1} \frac{x e ^{x}}{( 1 + x ) ^{2}} d x

Apply Integration By Parts

: [- \frac{e ^{x} x}{1 + x} - \int - e^{x} d x]_{0}^{1}

= [- \frac{e ^{x} x}{1 + x} - \int - e^{x} d x]_{0}^{1}

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

= [- \frac{e ^{x} x}{1 + x} - (- e^{x})]_{0}^{1}

Simplify

[- \frac{e ^{x} x}{1 + x} - (- e^{x})]_{0}^{1} : [\frac{e ^{x}}{1 + x}]_{0}^{1}

= [\frac{e ^{x}}{1 + x}]_{0}^{1}

Compute the boundaries:

\frac{e}{2} - 1

= \frac{e}{2} - 1

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What is the integral from 0 to 1 of (xe^x)/((1+x)^2) ?
The integral from 0 to 1 of (xe^x)/((1+x)^2) is e/2-1