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

The integral of (x-1/(x^2))(x^2+2/x)^3 is (x^8)/8+x^5+3x^2+4/x+2/(x^4)+C

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

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

\frac{x ^{8}}{8} + x^{5} + 3 x^{2} + \frac{4}{x} + \frac{2}{x ^{4}} + C

Solution steps

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

Expand

(x - \frac{1}{x ^{2}}) (x^{2} + \frac{2}{x})^{3} : x^{7} + 5 x^{4} + 6 x - \frac{4}{x ^{2}} - \frac{8}{x ^{5}}

= \int x^{7} + 5 x^{4} + 6 x - \frac{4}{x ^{2}} - \frac{8}{x ^{5}} 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^{7} d x + \int 5 x^{4} d x + \int 6 x d x - \int \frac{4}{x ^{2}} d x - \int \frac{8}{x ^{5}} d x

\int x^{7} d x = \frac{x ^{8}}{8}

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

\int 6 x d x = 3 x^{2}

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

\int \frac{8}{x ^{5}} d x = - \frac{2}{x ^{4}}

= \frac{x ^{8}}{8} + x^{5} + 3 x^{2} - (- \frac{4}{x}) - (- \frac{2}{x ^{4}})

Simplify

= \frac{x ^{8}}{8} + x^{5} + 3 x^{2} + \frac{4}{x} + \frac{2}{x ^{4}}

Add a constant to the solution

= \frac{x ^{8}}{8} + x^{5} + 3 x^{2} + \frac{4}{x} + \frac{2}{x ^{4}} + C

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What is the integral of (x-1/(x^2))(x^2+2/x)^3 ?
The integral of (x-1/(x^2))(x^2+2/x)^3 is (x^8)/8+x^5+3x^2+4/x+2/(x^4)+C