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

The integral of (x^3-2x)^2(1/x-5) is (x^6)/6-(5x^7)/7-x^4+4x^5+2x^2-(20x^3)/3+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{x ^{6}}{6} - \frac{5 x ^{7}}{7} - x^{4} + 4 x^{5} + 2 x^{2} - \frac{20 x ^{3}}{3} + 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) : x^{5} - 5 x^{6} - 4 x^{3} + 20 x^{4} + 4 x - 20 x^{2}

= \int x^{5} - 5 x^{6} - 4 x^{3} + 20 x^{4} + 4 x - 20 x^{2} 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^{5} d x - \int 5 x^{6} d x - \int 4 x^{3} d x + \int 20 x^{4} d x + \int 4 x d x - \int 20 x^{2} d x

\int x^{5} d x = \frac{x ^{6}}{6}

\int 5 x^{6} d x = \frac{5 x ^{7}}{7}

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

\int 20 x^{4} d x = 4 x^{5}

\int 4 x d x = 2 x^{2}

\int 20 x^{2} d x = \frac{20 x ^{3}}{3}

= \frac{x ^{6}}{6} - \frac{5 x ^{7}}{7} - x^{4} + 4 x^{5} + 2 x^{2} - \frac{20 x ^{3}}{3}

Add a constant to the solution

= \frac{x ^{6}}{6} - \frac{5 x ^{7}}{7} - x^{4} + 4 x^{5} + 2 x^{2} - \frac{20 x ^{3}}{3} + C

tan (e^{8 t}) + e^{t a n (8 t)}

f (x) = \frac{x}{e ^{x}}, at x = - 1

y^{^{'}} = 8 x e^{(5 y)}, y (0) = 0

\frac{d}{d x} (arccot (2 x))

\int \frac{7 ( x ^{2} + 5 )}{x ( x ^{2} + 7 )} d x

What is the integral of (x^3-2x)^2(1/x-5) ?
The integral of (x^3-2x)^2(1/x-5) is (x^6)/6-(5x^7)/7-x^4+4x^5+2x^2-(20x^3)/3+C