What is the integral of (1-x)e^{1+2x} ?

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

\int (1 - x) e^{1 + 2 x} d x

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

Solution steps

\int (1 - x) e^{1 + 2 x} d x

(1 - x) e^{1 + 2 x} = e^{1} (1 - x) e^{2 x}

= \int e^{1} (1 - x) e^{2 x} d x

Take the constant out:

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

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

Simplify

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

Apply Integration By Parts

: \frac{1}{2} e^{2 x} (1 - x) - \int - \frac{1}{2} e^{2 x} d x

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

\int - \frac{1}{2} e^{2 x} d x = - \frac{1}{4} e^{2 x}

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

Simplify

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

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

Add a constant to the solution

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

\int \frac{x ^{3} ( x ^{3} + 2 )}{15} d x

\int \frac{4 x ^{3}}{4 - x ^{2}} d x

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

\int x (x^{2} - 4)^{8} d x

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

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