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

The integral of (x^3+4)^4(3x^2) is 1/5 (x^3+4)^5+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

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

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

\frac{1}{5} (x^{3} + 4)^{5} + C

Solution steps

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

Apply u-substitution

: \int \frac{u ^{4}}{3} d u

= 3 \cdot \int \frac{u ^{4}}{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^{4} d u

Apply the Power Rule

: \frac{u ^{5}}{5}

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

Substitute back

u = x^{3} + 4

= 3 \cdot \frac{1}{3} \cdot \frac{( x ^{3} + 4 ) ^{5}}{5}

Simplify

3 \cdot \frac{1}{3} \cdot \frac{( x ^{3} + 4 ) ^{5}}{5} : \frac{1}{5} (x^{3} + 4)^{5}

= \frac{1}{5} (x^{3} + 4)^{5}

Add a constant to the solution

= \frac{1}{5} (x^{3} + 4)^{5} + C

\frac{d}{d x} (x x)

\int 3 d x

x \to 0 - lim (e^{\frac{8}{x}})

x + 1 - x^{2}

\int cot^{5} (6 x) d x

What is the integral of (x^3+4)^4(3x^2) ?
The integral of (x^3+4)^4(3x^2) is 1/5 (x^3+4)^5+C