What is the integral of 1/2 x^2*1/x ?

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

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integral of 1/2 x^2*1/x

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

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

Solution steps

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

= \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

x \to \infty lim ((\frac{x ^{p}}{e ^{x}})^{\frac{1}{n}})

\int sin (nπ t) d t

x \cdot \frac{d y}{d x} + 6 y = x^{3} - x

\frac{d}{d t} (4)

(\frac{2 t}{t ^{2} + 1})^{^{'}}