What is the integral of xe^{9x^2} ?

The integral of xe^{9x^2} is 1/18 e^{9x^2}+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of xe^{9x^2}

\int x e^{9 x^{2}} d x

\frac{1}{18} e^{9 x^{2}} + C

Solution steps

\int x e^{9 x^{2}} d x

Apply u-substitution

: \int \frac{e ^{u}}{18} d u

= \int \frac{e ^{u}}{18} d u

Take the constant out:

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

= \frac{1}{18} \cdot \int e^{u} d u

Use the common integral:

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

= \frac{1}{18} e^{u}

Substitute back

u = 9 x^{2}

= \frac{1}{18} e^{9 x^{2}}

Add a constant to the solution

= \frac{1}{18} e^{9 x^{2}} + C

\frac{d}{d x} (\frac{48}{x})

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

f (x) = 5 x - 4 x^{2}, (- 3, - 51)

\int_{0}^{1} x^{n} d x

\frac{4}{e ^{x} + e ^{- x}}

What is the integral of xe^{9x^2} ?
The integral of xe^{9x^2} is 1/18 e^{9x^2}+C