What is the integral of (4x^3-2/(x^2)-1) ?

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

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integral of (4x^3-2/(x^2)-1)

\int (4 x^{3} - \frac{2}{x ^{2}} - 1) d x

x^{4} + \frac{2}{x} - x + C

Solution steps

\int 4 x^{3} - \frac{2}{x ^{2}} - 1 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 4 x^{3} d x - \int \frac{2}{x ^{2}} d x - \int 1 d x

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

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

\int 1 d x = x

= x^{4} - (- \frac{2}{x}) - x

Simplify

= x^{4} + \frac{2}{x} - x

Add a constant to the solution

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

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