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

The integral of x(3x^2-1)^3 is 1/24 (3x^2-1)^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(3x^2-1)^3

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

\frac{1}{24} (3 x^{2} - 1)^{4} + C

Solution steps

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

Apply u-substitution

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

= \int \frac{u ^{3}}{6} d u

Take the constant out:

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

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

Apply the Power Rule

: \frac{u ^{4}}{4}

= \frac{1}{6} \cdot \frac{u ^{4}}{4}

Substitute back

u = 3 x^{2} - 1

= \frac{1}{6} \cdot \frac{( 3 x ^{2} - 1 ) ^{4}}{4}

Simplify

\frac{1}{6} \cdot \frac{( 3 x ^{2} - 1 ) ^{4}}{4} : \frac{1}{24} (3 x^{2} - 1)^{4}

= \frac{1}{24} (3 x^{2} - 1)^{4}

Add a constant to the solution

= \frac{1}{24} (3 x^{2} - 1)^{4} + C

\int \frac{12 x ^{2} + 16 x + 3}{4 x ^{3} + 8 x ^{2} + 3 x} d x

\int x^{3} x - 3 d x

\int x^{3} (x^{4} + 3)^{5} d x

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

\int \frac{1}{( x + 3 ) ( x + 4 )} d x

What is the integral of x(3x^2-1)^3 ?
The integral of x(3x^2-1)^3 is 1/24 (3x^2-1)^4+C