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

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

Popular >

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

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

\frac{e}{2} - 1

Decimal

0.35914 \dots

Solution steps

\int_{0}^{1} \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}^{1}

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

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

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

Simplify

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

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

Compute the boundaries:

\frac{e}{2} - 1

= \frac{e}{2} - 1

\int_{1}^{e} \frac{x}{x + 1} d x

\int_{0}^{21 π} sin (x) d x

\int_{0}^{x^{2}} cos (x) d x

\int_{0}^{2 π} - 150 sin^{2} (t) d t

\int_{1}^{16} x^{- \frac{5}{4}} d x

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