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

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

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

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

x + ln ∣ x ∣ + C

Solution steps

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

Expand

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

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

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

\int \frac{1}{x} d x = ln ∣ x ∣

= x + ln ∣ x ∣

Add a constant to the solution

= x + ln ∣ x ∣ + C

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