What is the integral of (6x^3+3/(x^2)-x) ?

The integral of (6x^3+3/(x^2)-x) is (3x^4)/2-3/x-(x^2)/2+C

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integral of (6x^3+3/(x^2)-x)

\int (6 x^{3} + \frac{3}{x ^{2}} - x) d x

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

Solution steps

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

\int 6 x^{3} d x = \frac{3 x ^{4}}{2}

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

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

= \frac{3 x ^{4}}{2} - \frac{3}{x} - \frac{x ^{2}}{2}

Add a constant to the solution

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

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

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