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

The integral of (x^2-2)/(2x) is (x^2)/4-ln|x|+C

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

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

\frac{x ^{2}}{4} - ln ∣ x ∣ + C

Solution steps

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

Expand

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

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

= \frac{1}{2} (\int x d x - \int \frac{2}{x} d x)

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

\int \frac{2}{x} d x = 2 ln ∣ x ∣

= \frac{1}{2} (\frac{x ^{2}}{2} - 2 ln ∣ x ∣)

Simplify

\frac{1}{2} (\frac{x ^{2}}{2} - 2 ln ∣ x ∣) : \frac{x ^{2}}{4} - ln ∣ x ∣

= \frac{x ^{2}}{4} - ln ∣ x ∣

Add a constant to the solution

= \frac{x ^{2}}{4} - ln ∣ x ∣ + C

\frac{\partial}{\partial x} (\frac{7 x - 5 y}{y ^{2} + 2 x ^{2}})

\frac{d}{d y} (3 x^{2} y)

\frac{d y}{d x} + 3 y = 0

\int_{1}^{e^{4}} x^{5} ln (x) d x

\frac{\partial}{\partial y} (y arctan (\frac{x}{z}))

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