What is the integral of x^2(4-2x^2) ?

The integral of x^2(4-2x^2) is 2(1/3 x^3(2-x^2)+(2x^5)/(15))+C

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integral of x^2(4-2x^2)

\int x^{2} (4 - 2 x^{2}) d x

2 (\frac{1}{3} x^{3} (2 - x^{2}) + \frac{2 x ^{5}}{15}) + C

Solution steps

\int x^{2} (4 - 2 x^{2}) d x

x^{2} (4 - 2 x^{2}) = 2 x^{2} (2 - x^{2})

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

Take the constant out:

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

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

Apply Integration By Parts

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

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

\int - \frac{2 x ^{4}}{3} d x = - \frac{2 x ^{5}}{15}

= 2 (\frac{1}{3} x^{3} (2 - x^{2}) - (- \frac{2 x ^{5}}{15}))

Simplify

= 2 (\frac{1}{3} x^{3} (2 - x^{2}) + \frac{2 x ^{5}}{15})

Add a constant to the solution

= 2 (\frac{1}{3} x^{3} (2 - x^{2}) + \frac{2 x ^{5}}{15}) + C

f (x) = x (x^{3} - x)

\frac{d}{d x} (\frac{6 x - 3}{5 x + 8})

\int \frac{4 x - 11}{2 x ^{2} - 11 x + 15} d x

\frac{\partial}{\partial y} (2 x + arctan (y^{2}))

\int e^{\frac{5}{2} x} d x

What is the integral of x^2(4-2x^2) ?
The integral of x^2(4-2x^2) is 2(1/3 x^3(2-x^2)+(2x^5)/(15))+C