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

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

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

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

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

Solution steps

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

Expand

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

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

Apply the Sum Rule:

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

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

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

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

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

Add a constant to the solution

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

y^{^{'}} = \frac{( 4 x sec ( \frac{y}{x} ) + y )}{x}

x \to - 6 lim (\frac{6}{x + 6})

y^{^{''}} + y = 8 x sin (x)

\frac{\partial}{\partial x} (x^{2} - 2 x y + 4 y^{2})

x \to 6 lim (6 \cdot \frac{5}{6} \cdot (- \frac{2}{3}))

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