What is the integral of (8+x^3)^5(3x^2) ?

The integral of (8+x^3)^5(3x^2) is 1/6 (8+x^3)^6+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of (8+x^3)^5(3x^2)

\int (8 + x^{3})^{5} (3 x^{2}) d x

\frac{1}{6} (8 + x^{3})^{6} + C

Solution steps

\int (8 + x^{3})^{5} \cdot 3 x^{2} d x

Take the constant out:

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

= 3 \cdot \int (8 + x^{3})^{5} x^{2} d x

Apply u-substitution

: \int \frac{u ^{5}}{3} d u

= 3 \cdot \int \frac{u ^{5}}{3} d u

Take the constant out:

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

= 3 \cdot \frac{1}{3} \cdot \int u^{5} d u

Apply the Power Rule

: \frac{u ^{6}}{6}

= 3 \cdot \frac{1}{3} \cdot \frac{u ^{6}}{6}

Substitute back

u = 8 + x^{3}

= 3 \cdot \frac{1}{3} \cdot \frac{( 8 + x ^{3} ) ^{6}}{6}

Simplify

3 \cdot \frac{1}{3} \cdot \frac{( 8 + x ^{3} ) ^{6}}{6} : \frac{1}{6} (8 + x^{3})^{6}

= \frac{1}{6} (8 + x^{3})^{6}

Add a constant to the solution

= \frac{1}{6} (8 + x^{3})^{6} + C

f (x) = \frac{1}{5 x ^{2} + 4 x}

\int_{1}^{2} (6 x^{2} - 3 x) d x

\frac{\partial}{\partial x} (x y - x)

x^{3} (x - 4)

\int (sin^{2} (x) cos^{2} (x)) d x

What is the integral of (8+x^3)^5(3x^2) ?
The integral of (8+x^3)^5(3x^2) is 1/6 (8+x^3)^6+C