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

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

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

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

\frac{x ^{4}}{4} - x^{2} + C

Solution steps

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

Expand

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

= \int x^{3} - 2 x d x

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= \int x^{3} d x - \int 2 x d x

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

\int 2 x d x = x^{2}

= \frac{x ^{4}}{4} - x^{2}

Add a constant to the solution

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

f (x) = arctan (\frac{1 + x}{1 - x})

\frac{d y}{d x} = 15 sin (\frac{π}{20} (x - 2)) + 2

y = tan (18 x^{2}), at x = π

\int 2 cos (7 x - 4) d x

y^{^{''}} + 9 y = csc^{2} (3 t)

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