What is the integral of t^2(t-8/t) ?

The integral of t^2(t-8/t) is (t^4)/4-4t^2+C

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integral of t^2(t-8/t)

\int t^{2} (t - \frac{8}{t}) d t

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

Solution steps

\int t^{2} (t - \frac{8}{t}) d t

Expand

t^{2} (t - \frac{8}{t}) : t^{3} - 8 t

= \int t^{3} - 8 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 8 t d t

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

\int 8 t d t = 4 t^{2}

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

Add a constant to the solution

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

\frac{d y}{d x} = (x e^{x^{2}})

y^{^{'}} - 2 y = y^{3}

\frac{d}{d x} (arccsc (tan (e^{x})))

\frac{\partial}{\partial x} (e^{2 x} cos (8 y))

(\frac{1}{x ( ln ( x ) ) ^{2}})^{^{'}}

What is the integral of t^2(t-8/t) ?
The integral of t^2(t-8/t) is (t^4)/4-4t^2+C