What is the integral of (1+x^3)^73x^2 ?

The integral of (1+x^3)^73x^2 is 1/8 (1+x^3)^8+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of (1+x^3)^73x^2

\int (1 + x^{3})^{7} 3 x^{2} d x

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

Solution steps

\int (1 + x^{3})^{7} \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 (1 + x^{3})^{7} x^{2} d x

Apply u-substitution

: \int \frac{u ^{7}}{3} d u

= 3 \cdot \int \frac{u ^{7}}{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^{7} d u

Apply the Power Rule

: \frac{u ^{8}}{8}

= 3 \cdot \frac{1}{3} \cdot \frac{u ^{8}}{8}

Substitute back

u = 1 + x^{3}

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

Simplify

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

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

Add a constant to the solution

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

\frac{\partial}{\partial y} (x^{2} e^{x y})

\int \frac{1 - x}{1 + x} d x

\frac{\partial}{\partial x} (\frac{sin ( x )}{r ^{2}})

\frac{d}{d x} (9 x^{l n (9 x)})

g (x) = - x^{5} + 4 x^{2} - 2 x + 2

What is the integral of (1+x^3)^73x^2 ?
The integral of (1+x^3)^73x^2 is 1/8 (1+x^3)^8+C