What is the integral of 6x^3-2x+1/(3x^2) ?

The integral of 6x^3-2x+1/(3x^2) is (3x^4)/2-x^2-1/(3x)+C

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integral of 6x^3-2x+1/(3x^2)

\int 6 x^{3} - 2 x + \frac{1}{3 x ^{2}} d x

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

Solution steps

\int 6 x^{3} - 2 x + \frac{1}{3 x ^{2}} 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 2 x d x + \int \frac{1}{3 x ^{2}} d x

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

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

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

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

Add a constant to the solution

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

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What is the integral of 6x^3-2x+1/(3x^2) ?
The integral of 6x^3-2x+1/(3x^2) is (3x^4)/2-x^2-1/(3x)+C