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

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

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

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

\frac{x ^{2}}{2} + 2 x + C

Solution steps

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

Expand

\frac{x ^{2} + x}{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 = 2 x

= \frac{x ^{2}}{2} + 2 x

Add a constant to the solution

= \frac{x ^{2}}{2} + 2 x + C

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