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

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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of (2x-1)e^{x(x-1)}

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

e^{x (x - 1)} + C

Solution steps

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

Apply u-substitution

: \int e^{u} d u

= \int e^{u} d u

Use the common integral:

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

= e^{u}

Substitute back

u = x (x - 1)

= e^{x (x - 1)}

Add a constant to the solution

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

x \to 1 + lim (x^{\frac{7}{1 - x}})

\int \frac{4 x ^{2}}{( 1 - x ^{2} ) ^{\frac{3}{2}}} d x

x y^{^{'}} + 3 y = x^{3} - x

\int x^{- 0.3} d x

\int \frac{82}{( x - 1 ) ( x ^{2} + 81 )} d x

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