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

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

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

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

\frac{1}{2} (\frac{x ^{2}}{2} + ln ∣ x ∣) + C

Solution steps

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

Expand

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

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

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

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

\int \frac{1}{x} d x = ln ∣ x ∣

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

Add a constant to the solution

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

x \to 5 - lim (\frac{∣ x - 5 ∣}{x - 5})

\frac{d}{d x} (\frac{x ^{2}}{7 x + 4})

\frac{\partial}{\partial y} (x^{2} \cdot y)

\frac{d}{d x} (\frac{4 x}{6 x - 1})

x \to 4 lim (2 x - 7)

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