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

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

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

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

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

Solution steps

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

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

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

\int \frac{1}{x} d x = ln ∣ x ∣

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

Add a constant to the solution

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

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