What is the integral of (x-1/(x^2))(x^2+2/x)^4 ?

The integral of (x-1/(x^2))(x^2+2/x)^4 is (x^{10})/(10)+x^7+4x^4+8/(x^2)+(16)/(5x^5)+8x+C

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integral of (x-1/(x^2))(x^2+2/x)^4

\int (x - \frac{1}{x ^{2}}) (x^{2} + \frac{2}{x})^{4} d x

\frac{x ^{10}}{10} + x^{7} + 4 x^{4} + \frac{8}{x ^{2}} + \frac{16}{5 x ^{5}} + 8 x + C

Solution steps

\int (x - \frac{1}{x ^{2}}) (x^{2} + \frac{2}{x})^{4} d x

Expand

(x - \frac{1}{x ^{2}}) (x^{2} + \frac{2}{x})^{4} : x^{9} + 7 x^{6} + 16 x^{3} - \frac{16}{x ^{3}} - \frac{16}{x ^{6}} + 8

= \int x^{9} + 7 x^{6} + 16 x^{3} - \frac{16}{x ^{3}} - \frac{16}{x ^{6}} + 8 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^{9} d x + \int 7 x^{6} d x + \int 16 x^{3} d x - \int \frac{16}{x ^{3}} d x - \int \frac{16}{x ^{6}} d x + \int 8 d x

\int x^{9} d x = \frac{x ^{10}}{10}

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

\int 16 x^{3} d x = 4 x^{4}

\int \frac{16}{x ^{3}} d x = - \frac{8}{x ^{2}}

\int \frac{16}{x ^{6}} d x = - \frac{16}{5 x ^{5}}

\int 8 d x = 8 x

= \frac{x ^{10}}{10} + x^{7} + 4 x^{4} - (- \frac{8}{x ^{2}}) - (- \frac{16}{5 x ^{5}}) + 8 x

Simplify

= \frac{x ^{10}}{10} + x^{7} + 4 x^{4} + \frac{8}{x ^{2}} + \frac{16}{5 x ^{5}} + 8 x

Add a constant to the solution

= \frac{x ^{10}}{10} + x^{7} + 4 x^{4} + \frac{8}{x ^{2}} + \frac{16}{5 x ^{5}} + 8 x + C

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What is the integral of (x-1/(x^2))(x^2+2/x)^4 ?
The integral of (x-1/(x^2))(x^2+2/x)^4 is (x^{10})/(10)+x^7+4x^4+8/(x^2)+(16)/(5x^5)+8x+C