What is the integral of 7x^2(6x^3+3)^3 ?

The integral of 7x^2(6x^3+3)^3 is 7/72 (6x^3+3)^4+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of 7x^2(6x^3+3)^3

\int 7 x^{2} (6 x^{3} + 3)^{3} d x

\frac{7}{72} (6 x^{3} + 3)^{4} + C

Solution steps

\int 7 x^{2} (6 x^{3} + 3)^{3} d x

Take the constant out:

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

= 7 \cdot \int x^{2} (6 x^{3} + 3)^{3} d x

Apply u-substitution

: \int \frac{u ( 6 u ^{\frac{3}{2}} + 3 ) ^{3}}{2 x} d u

= 7 \cdot \int \frac{( 6 u + 3 ) ^{3}}{3} d u

Take the constant out:

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

= 7 \cdot \frac{1}{3} \cdot \int (6 u + 3)^{3} d u

Apply u-substitution

: \int \frac{v ^{3}}{6} d v

= 7 \cdot \frac{1}{3} \cdot \int \frac{v ^{3}}{6} d v

Take the constant out:

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

= 7 \cdot \frac{1}{3} \cdot \frac{1}{6} \cdot \int v^{3} d v

Apply the Power Rule

: \frac{v ^{4}}{4}

= 7 \cdot \frac{1}{3} \cdot \frac{1}{6} \cdot \frac{v ^{4}}{4}

Substitute back

= 7 \cdot \frac{1}{3} \cdot \frac{1}{6} \cdot \frac{( 6 x ^{3} + 3 ) ^{4}}{4}

Simplify

7 \cdot \frac{1}{3} \cdot \frac{1}{6} \cdot \frac{( 6 x ^{3} + 3 ) ^{4}}{4} : \frac{7}{72} (6 x^{3} + 3)^{4}

= \frac{7}{72} (6 x^{3} + 3)^{4}

Add a constant to the solution

= \frac{7}{72} (6 x^{3} + 3)^{4} + C

\int \frac{x ^{2}}{x - 4} d x

\frac{\partial}{\partial x} (4 x^{3} y - y sin (x))

\int \frac{1}{25 x ^{2} - 16} d x

x \to 0 + lim (x ln (3 x))

\frac{\partial}{\partial z} (- 2 x y)

What is the integral of 7x^2(6x^3+3)^3 ?
The integral of 7x^2(6x^3+3)^3 is 7/72 (6x^3+3)^4+C