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

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

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

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

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

Solution steps

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

Expand

\frac{6 x ^{3} + 1}{x ^{2}} : 6 x + \frac{1}{x ^{2}}

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

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

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

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

Add a constant to the solution

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

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