What is the integral of 32sqrt(x)e^{sqrt(x)} ?

The integral of 32sqrt(x)e^{sqrt(x)} is 64(e^{sqrt(x)}x-2(e^{sqrt(x)}sqrt(x)-e^{sqrt(x)}))+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of 32sqrt(x)e^{sqrt(x)}

\int 32 x e^{x} d x

64 (e^{x} x - 2 (e^{x} x - e^{x})) + C

Solution steps

\int 32 x e^{x} d x

Take the constant out:

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

= 32 \cdot \int x e^{x} d x

Apply u-substitution

: \int 2 e^{u} u^{2} d u

= 32 \cdot \int 2 e^{u} u^{2} d u

Take the constant out:

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

= 32 \cdot 2 \cdot \int e^{u} u^{2} d u

Apply Integration By Parts

: u^{2} e^{u} - \int 2 u e^{u} d u

= 32 \cdot 2 (u^{2} e^{u} - \int 2 u e^{u} d u)

\int 2 u e^{u} d u = 2 (e^{u} u - e^{u})

= 32 \cdot 2 (u^{2} e^{u} - 2 (e^{u} u - e^{u}))

Substitute back

u = x

= 32 \cdot 2 ((x)^{2} e^{x} - 2 (e^{x} x - e^{x}))

Simplify

32 \cdot 2 ((x)^{2} e^{x} - 2 (e^{x} x - e^{x})) : 64 (e^{x} x - 2 (e^{x} x - e^{x}))

= 64 (e^{x} x - 2 (e^{x} x - e^{x}))

Add a constant to the solution

= 64 (e^{x} x - 2 (e^{x} x - e^{x})) + C

\int \frac{8 x}{81 - 16 x ^{4}} d x

\int \frac{x ln ^{2} ( 1 + x ^{2} )}{1 + x ^{2}} d x

\int \frac{5 x + 1}{( x + 1 ) ( x - 1 )} d x

\int (x^{3} (2 + x^{4})^{5}) d x

\int (2 x e^{x} - y + 6 x^{2}) d x

What is the integral of 32sqrt(x)e^{sqrt(x)} ?
The integral of 32sqrt(x)e^{sqrt(x)} is 64(e^{sqrt(x)}x-2(e^{sqrt(x)}sqrt(x)-e^{sqrt(x)}))+C