What is the integral of (e^x)^2x ?

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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of (e^x)^2x

integral

(e^{x})^{2} x

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

Solution steps

\int (e^{x})^{2} x d x

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

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

Apply u-substitution

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

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

Take the constant out:

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

= \frac{1}{4} \cdot \int e^{u} u d u

Apply Integration By Parts

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

= \frac{1}{4} (e^{u} u - \int e^{u} d u)

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

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

Substitute back

u = 2 x

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

Add a constant to the solution

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

\frac{1}{( 5 t + 2 )}

1 + e^{v}

y^{2} ((1 - y^{2})^{\frac{1}{2}})

x^{4} + 2 x + 3

\frac{( e ^{x} )}{x ^{- 1}}

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