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

The integral of (2x-1/(x^2))^3 is 2x^4-12x-3/(x^2)+1/(5x^5)+C

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

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

2 x^{4} - 12 x - \frac{3}{x ^{2}} + \frac{1}{5 x ^{5}} + C

Solution steps

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

Expand

(2 x - \frac{1}{x ^{2}})^{3} : 8 x^{3} - 12 + \frac{6}{x ^{3}} - \frac{1}{x ^{6}}

= \int 8 x^{3} - 12 + \frac{6}{x ^{3}} - \frac{1}{x ^{6}} 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 8 x^{3} d x - \int 12 d x + \int \frac{6}{x ^{3}} d x - \int \frac{1}{x ^{6}} d x

\int 8 x^{3} d x = 2 x^{4}

\int 12 d x = 12 x

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

\int \frac{1}{x ^{6}} d x = - \frac{1}{5 x ^{5}}

= 2 x^{4} - 12 x - \frac{3}{x ^{2}} - (- \frac{1}{5 x ^{5}})

Simplify

= 2 x^{4} - 12 x - \frac{3}{x ^{2}} + \frac{1}{5 x ^{5}}

Add a constant to the solution

= 2 x^{4} - 12 x - \frac{3}{x ^{2}} + \frac{1}{5 x ^{5}} + C

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What is the integral of (2x-1/(x^2))^3 ?
The integral of (2x-1/(x^2))^3 is 2x^4-12x-3/(x^2)+1/(5x^5)+C