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

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

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

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

\frac{3 x ^{2}}{2} + 6 x + ln ∣ x ∣ + C

Solution steps

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

Expand

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

= \int 3 x + 6 + \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 3 x d x + \int 6 d x + \int \frac{1}{x} d x

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

\int 6 d x = 6 x

\int \frac{1}{x} d x = ln ∣ x ∣

= \frac{3 x ^{2}}{2} + 6 x + ln ∣ x ∣

Add a constant to the solution

= \frac{3 x ^{2}}{2} + 6 x + ln ∣ x ∣ + C

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