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

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

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

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

3 x^{2} + ln ∣ x ∣ + C

Solution steps

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

Expand

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

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

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

\int \frac{1}{x} d x = ln ∣ x ∣

= 3 x^{2} + ln ∣ x ∣

Add a constant to the solution

= 3 x^{2} + ln ∣ x ∣ + C