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

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

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

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

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

Solution steps

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

Apply Integration By Parts

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

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

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

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

Simplify

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

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

Add a constant to the solution

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

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

y^{^{'''}} + 4 y^{^{''}} = 3 x - 1

\int e^{(x + 3)} d x

\int x^{2} cos (9 x) d x

\frac{1}{5 s}

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