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

The integral of (xe^x)/(x+1) is 1/e (e^{x+1}-\Ei(x+1))+C

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

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

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

Solution steps

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

Apply u-substitution

: \int \frac{e ^{u - 1} ( u - 1 )}{u} d u

= \int \frac{e ^{u - 1} ( u - 1 )}{u} d u

Simplify

e^{u - 1} : e^{u} e^{- 1}

= \int \frac{e ^{u} e ^{- 1} ( u - 1 )}{u} d u

Take the constant out:

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

= e^{- 1} \cdot \int \frac{e ^{u} ( u - 1 )}{u} d u

Apply exponent rule:

a^{- 1} = \frac{1}{a}

= \frac{1}{e} \cdot \int \frac{e ^{u} ( u - 1 )}{u} d u

Expand

\frac{e ^{u} ( u - 1 )}{u} : e^{u} - \frac{e ^{u}}{u}

= \frac{1}{e} \cdot \int e^{u} - \frac{e ^{u}}{u} d u

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= \frac{1}{e} (\int e^{u} d u - \int \frac{e ^{u}}{u} d u)

\int e^{u} d u = e^{u}

\int \frac{e ^{u}}{u} d u = Ei (u)

= \frac{1}{e} (e^{u} - Ei (u))

Substitute back

u = x + 1

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

Add a constant to the solution

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

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What is the integral of (xe^x)/(x+1) ?
The integral of (xe^x)/(x+1) is 1/e (e^{x+1}-\Ei(x+1))+C