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

The integral of (e^{2x}-2x)^2 is 1/4 e^{4x}-2e^{2x}x+e^{2x}+(4x^3)/3+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of (e^{2x}-2x)^2

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

\frac{1}{4} e^{4 x} - 2 e^{2 x} x + e^{2 x} + \frac{4 x ^{3}}{3} + C

Solution steps

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

Apply u-substitution

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

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

Take the constant out:

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

= \frac{1}{2} \cdot \int (e^{u} - u)^{2} d u

Expand

(e^{u} - u)^{2} : e^{2 u} - 2 e^{u} u + u^{2}

= \frac{1}{2} \cdot \int e^{2 u} - 2 e^{u} u + u^{2} 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}{2} (\int e^{2 u} d u - \int 2 e^{u} u d u + \int u^{2} d u)

\int e^{2 u} d u = \frac{1}{2} e^{2 u}

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

\int u^{2} d u = \frac{u ^{3}}{3}

= \frac{1}{2} (\frac{1}{2} e^{2 u} - 2 (e^{u} u - e^{u}) + \frac{u ^{3}}{3})

Substitute back

u = 2 x

= \frac{1}{2} (\frac{1}{2} e^{2 \cdot 2 x} - 2 (e^{2 x} \cdot 2 x - e^{2 x}) + \frac{( 2 x ) ^{3}}{3})

Simplify

\frac{1}{2} (\frac{1}{2} e^{2 \cdot 2 x} - 2 (e^{2 x} \cdot 2 x - e^{2 x}) + \frac{( 2 x ) ^{3}}{3}) : \frac{1}{4} e^{4 x} - 2 e^{2 x} x + e^{2 x} + \frac{4 x ^{3}}{3}

= \frac{1}{4} e^{4 x} - 2 e^{2 x} x + e^{2 x} + \frac{4 x ^{3}}{3}

Add a constant to the solution

= \frac{1}{4} e^{4 x} - 2 e^{2 x} x + e^{2 x} + \frac{4 x ^{3}}{3} + C

\int 2^{0.1 x^{2}} d x

\int (cos (x))^{7} d x

\int \frac{sin ( 1 - x )}{1 - x} d x

\int y^{2} e^{s i n (2 x)}

\int 0.0003 x^{2} - 0.12 x + 20 d x

What is the integral of (e^{2x}-2x)^2 ?
The integral of (e^{2x}-2x)^2 is 1/4 e^{4x}-2e^{2x}x+e^{2x}+(4x^3)/3+C