What is the integral of xe^{4x^2} ?

The integral of xe^{4x^2} is 1/8 e^{4x^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^{4x^2}

\int x e^{4 x^{2}} d x

\frac{1}{8} e^{4 x^{2}} + C

Solution steps

\int x e^{4 x^{2}} d x

Apply u-substitution

: \int \frac{e ^{u}}{8} d u

= \int \frac{e ^{u}}{8} d u

Take the constant out:

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

= \frac{1}{8} \cdot \int e^{u} d u

Use the common integral:

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

= \frac{1}{8} e^{u}

Substitute back

u = 4 x^{2}

= \frac{1}{8} e^{4 x^{2}}

Add a constant to the solution

= \frac{1}{8} e^{4 x^{2}} + C

x \to 5 + lim (\frac{x + 5}{x + 5})

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

n = 1 \sum \infty \frac{1}{2 \cdot n}

\frac{x}{x + 5}

\int \frac{4}{64 - 36 x ^{2}} d x