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

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

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integral of (x^2+1/x)

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

\frac{x ^{3}}{3} + ln ∣ x ∣ + C

Solution steps

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

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

\int \frac{1}{x} d x = ln ∣ x ∣

= \frac{x ^{3}}{3} + ln ∣ x ∣

Add a constant to the solution

= \frac{x ^{3}}{3} + ln ∣ x ∣ + C

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