What is the integral of (x^3-2x^2)(1/x-5) ?

The integral of (x^3-2x^2)(1/x-5) is -(5x^4)/4+(11x^3)/3-x^2+C

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integral of (x^3-2x^2)(1/x-5)

\int (x^{3} - 2 x^{2}) (\frac{1}{x} - 5) d x

- \frac{5 x ^{4}}{4} + \frac{11 x ^{3}}{3} - x^{2} + C

Solution steps

\int (x^{3} - 2 x^{2}) (\frac{1}{x} - 5) d x

Expand

(x^{3} - 2 x^{2}) (\frac{1}{x} - 5) : - 5 x^{3} + 11 x^{2} - 2 x

= \int - 5 x^{3} + 11 x^{2} - 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 5 x^{3} d x + \int 11 x^{2} d x - \int 2 x d x

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

\int 11 x^{2} d x = \frac{11 x ^{3}}{3}

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

= - \frac{5 x ^{4}}{4} + \frac{11 x ^{3}}{3} - x^{2}

Add a constant to the solution

= - \frac{5 x ^{4}}{4} + \frac{11 x ^{3}}{3} - x^{2} + C

x \to 2 - lim (ln (2 - x))

\int x e^{8 x^{2}} d x

\frac{d y}{d t} = e^{t} (1 - 1)^{2}

- 5 t^{2}

f (x) = \frac{( x ^{2} + 1 )}{x - 1}

What is the integral of (x^3-2x^2)(1/x-5) ?
The integral of (x^3-2x^2)(1/x-5) is -(5x^4)/4+(11x^3)/3-x^2+C