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

The integral of ((x^2)/2)(1/x) is (x^2)/4+C

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

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

\frac{x ^{2}}{4} + C

Solution steps

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

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

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

Simplify

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

= \frac{1}{2} \cdot \int x d x

Apply the Power Rule

= \frac{1}{2} \cdot \frac{x ^{2}}{2}

Simplify

\frac{1}{2} \cdot \frac{x ^{2}}{2} : \frac{x ^{2}}{4}

= \frac{x ^{2}}{4}

Add a constant to the solution

= \frac{x ^{2}}{4} + C

What is the integral of ((x^2)/2)(1/x) ?
The integral of ((x^2)/2)(1/x) is (x^2)/4+C