What is the integral of (xe^x)/(sqrt(e^{x-1))} ?

The integral of (xe^x)/(sqrt(e^{x-1))} is 4sqrt(e)(1/2 e^{x/2}x-e^{x/2})+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)/(sqrt(e^{x-1))}

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

4 e (\frac{1}{2} e^{\frac{x}{2}} x - e^{\frac{x}{2}}) + C

Solution steps

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

\frac{x e ^{x}}{e ^{x - 1}} = \frac{1}{e ^{- 1}} \frac{x e ^{x}}{e ^{x}}

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

Take the constant out:

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

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

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

= e \cdot \int \frac{x e ^{x}}{e ^{x}} d x

Simplify

\frac{x e ^{x}}{e ^{x}} : e^{\frac{x}{2}} x

= e \cdot \int e^{\frac{x}{2}} x d x

Apply u-substitution

: \int 4 e^{u} u d u

= e \cdot \int 4 e^{u} u d u

Take the constant out:

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

= e \cdot 4 \cdot \int e^{u} u d u

Apply Integration By Parts

: e^{u} u - \int e^{u} d u

= e \cdot 4 (e^{u} u - \int e^{u} d u)

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

= e \cdot 4 (e^{u} u - e^{u})

Substitute back

u = \frac{x}{2}

= e \cdot 4 (e^{\frac{x}{2}} \frac{x}{2} - e^{\frac{x}{2}})

Simplify

e \cdot 4 (e^{\frac{x}{2}} \frac{x}{2} - e^{\frac{x}{2}}) : 4 e (\frac{1}{2} e^{\frac{x}{2}} x - e^{\frac{x}{2}})

= 4 e (\frac{1}{2} e^{\frac{x}{2}} x - e^{\frac{x}{2}})

Add a constant to the solution

= 4 e (\frac{1}{2} e^{\frac{x}{2}} x - e^{\frac{x}{2}}) + C

\int e^{x^{2} + 3 x} d x

\int sin (t) d t

\int \frac{( 2 - x )}{x ^{2}} d x

\int (3 x + 1)^{6} d x

\int 6 x^{- \frac{6}{7}} d x

What is the integral of (xe^x)/(sqrt(e^{x-1))} ?
The integral of (xe^x)/(sqrt(e^{x-1))} is 4sqrt(e)(1/2 e^{x/2}x-e^{x/2})+C