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

The integral of 3x(x^2-1/3) is 3(1/6 x^2(3x^2-1)-(x^4)/4)+C

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integral of 3x(x^2-1/3)

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

3 (\frac{1}{6} x^{2} (3 x^{2} - 1) - \frac{x ^{4}}{4}) + C

Solution steps

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

Take the constant out:

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

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

Apply Integration By Parts

: \frac{1}{6} x^{2} (3 x^{2} - 1) - \int x^{3} d x

= 3 (\frac{1}{6} x^{2} (3 x^{2} - 1) - \int x^{3} d x)

\int x^{3} d x = \frac{x ^{4}}{4}

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

Add a constant to the solution

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

f (x) = \frac{x}{x + 1}, at x = - 2

\frac{\partial}{\partial x} (- \frac{1}{2 x ^{2}})

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

\frac{d}{d x} (sin^{- 1} (x))

\int sec^{2} (X) d X

What is the integral of 3x(x^2-1/3) ?
The integral of 3x(x^2-1/3) is 3(1/6 x^2(3x^2-1)-(x^4)/4)+C