What is the integral of x^2(x^3+4)^2 ?

The integral of x^2(x^3+4)^2 is 1/9 (x^3+4)^3+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

integral of x^2(x^3+4)^2

\int x^{2} (x^{3} + 4)^{2} d x

\frac{1}{9} (x^{3} + 4)^{3} + C

Solution steps

\int x^{2} (x^{3} + 4)^{2} d x

Apply u-substitution

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

Apply the Power Rule

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

Substitute back

u = x^{3} + 4

= \frac{1}{3} \cdot \frac{( x ^{3} + 4 ) ^{3}}{3}

Simplify

\frac{1}{3} \cdot \frac{( x ^{3} + 4 ) ^{3}}{3} : \frac{1}{9} (x^{3} + 4)^{3}

= \frac{1}{9} (x^{3} + 4)^{3}

Add a constant to the solution

= \frac{1}{9} (x^{3} + 4)^{3} + C

What is the integral of x^2(x^3+4)^2 ?
The integral of x^2(x^3+4)^2 is 1/9 (x^3+4)^3+C