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

The integral of x^2(3x^3+7)^3 is 1/36 (3x^3+7)^4+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

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

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

\frac{1}{36} (3 x^{3} + 7)^{4} + C

Solution steps

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

Apply u-substitution

: \int \frac{u ( 3 u ^{\frac{3}{2}} + 7 ) ^{3}}{2 x} d u

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

Apply u-substitution

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

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

Take the constant out:

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

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

Apply the Power Rule

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

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

Substitute back

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

Simplify

\frac{1}{3} \cdot \frac{1}{3} \cdot \frac{( 3 x ^{3} + 7 ) ^{4}}{4} : \frac{1}{36} (3 x^{3} + 7)^{4}

= \frac{1}{36} (3 x^{3} + 7)^{4}

Add a constant to the solution

= \frac{1}{36} (3 x^{3} + 7)^{4} + C

\int (tan (x) sec^{2} (x)) d x

sin (\frac{1}{2} t) \cdot t^{2}

\int - cos (x) 2 x d x

y^{^{'}} = 4 y

\frac{d}{d x} (\frac{x ^{3} + 2 x ^{2}}{x})

What is the integral of x^2(3x^3+7)^3 ?
The integral of x^2(3x^3+7)^3 is 1/36 (3x^3+7)^4+C