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

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

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

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

\frac{x ^{2}}{2} + ln ∣ x ∣ + C

Solution steps

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

Expand

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

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

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

\int \frac{1}{x} d x = ln ∣ x ∣

= \frac{x ^{2}}{2} + ln ∣ x ∣

Add a constant to the solution

= \frac{x ^{2}}{2} + ln ∣ x ∣ + C

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