What is the integral of (x^3)/6+1/(2x) ?

The integral of (x^3)/6+1/(2x) is 1/6 ((x^4)/4+3ln|x|)+C

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integral of (x^3)/6+1/(2x)

\int \frac{x ^{3}}{6} + \frac{1}{2 x} d x

\frac{1}{6} (\frac{x ^{4}}{4} + 3 ln ∣ x ∣) + C

Solution steps

\int \frac{x ^{3}}{6} + \frac{1}{2 x} d x

Simplify

\frac{x ^{3}}{6} + \frac{1}{2 x} : \frac{x ^{4} + 3}{6 x}

= \int \frac{x ^{4} + 3}{6 x} d x

Take the constant out:

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

= \frac{1}{6} \cdot \int \frac{x ^{4} + 3}{x} d x

Expand

\frac{x ^{4} + 3}{x} : x^{3} + \frac{3}{x}

= \frac{1}{6} \cdot \int x^{3} + \frac{3}{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}{6} (\int x^{3} d x + \int \frac{3}{x} d x)

\int x^{3} d x = \frac{x ^{4}}{4}

\int \frac{3}{x} d x = 3 ln ∣ x ∣

= \frac{1}{6} (\frac{x ^{4}}{4} + 3 ln ∣ x ∣)

Add a constant to the solution

= \frac{1}{6} (\frac{x ^{4}}{4} + 3 ln ∣ x ∣) + C

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What is the integral of (x^3)/6+1/(2x) ?
The integral of (x^3)/6+1/(2x) is 1/6 ((x^4)/4+3ln|x|)+C