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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

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

\int x^{2} - 4 x - 45 d x

\int 2 + cos (θ) d θ

\int (2 x - 5) (3 x + 2) d x

\int (e^{- 3 t} + cos (2 t)) d t

\int \frac{6 x ^{3} + 1}{x ^{2}} d x

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