What is the integral of x^2e^{x^3+8} ?

The integral of x^2e^{x^3+8} is 1/3 e^{x^3+8}+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

integral of x^2e^{x^3+8}

\int x^{2} e^{x^{3} + 8} d x

\frac{1}{3} e^{x^{3} + 8} + C

Solution steps

\int x^{2} e^{x^{3} + 8} d x

Apply u-substitution

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

Take the constant out:

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

= \frac{1}{3} \cdot \int e^{u} d u

Use the common integral:

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

= \frac{1}{3} e^{u}

Substitute back

u = x^{3} + 8

= \frac{1}{3} e^{x^{3} + 8}

Add a constant to the solution

= \frac{1}{3} e^{x^{3} + 8} + C

What is the integral of x^2e^{x^3+8} ?
The integral of x^2e^{x^3+8} is 1/3 e^{x^3+8}+C