What is the integral from 0 to 3 of (xe^x)/((x+1)^2) ?

The integral from 0 to 3 of (xe^x)/((x+1)^2) is (e^3)/4-1

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integral from 0 to 3 of (xe^x)/((x+1)^2)

\int_{0}^{3} \frac{x e ^{x}}{( x + 1 ) ^{2}} d x

\frac{e ^{3}}{4} - 1

Decimal

4.02138 \dots

Solution steps

\int_{0}^{3} \frac{x e ^{x}}{( x + 1 ) ^{2}} d x

Apply Integration By Parts

: [- \frac{e ^{x} x}{x + 1} - \int - e^{x} d x]_{0}^{3}

= [- \frac{e ^{x} x}{x + 1} - \int - e^{x} d x]_{0}^{3}

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

= [- \frac{e ^{x} x}{x + 1} - (- e^{x})]_{0}^{3}

Simplify

[- \frac{e ^{x} x}{x + 1} - (- e^{x})]_{0}^{3} : [\frac{e ^{x}}{x + 1}]_{0}^{3}

= [\frac{e ^{x}}{x + 1}]_{0}^{3}

Compute the boundaries:

\frac{e ^{3}}{4} - 1

= \frac{e ^{3}}{4} - 1

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What is the integral from 0 to 3 of (xe^x)/((x+1)^2) ?
The integral from 0 to 3 of (xe^x)/((x+1)^2) is (e^3)/4-1