What is the integral of (x^3)/3+1/(4x) ?

The integral of (x^3)/3+1/(4x) is (x^4)/(12)+1/4 ln|x|+C

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integral of (x^3)/3+1/(4x)

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

\frac{x ^{4}}{12} + \frac{1}{4} ln ∣ x ∣ + C

Solution steps

\int \frac{x ^{3}}{3} + \frac{1}{4 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

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

\int \frac{x ^{3}}{3} d x = \frac{x ^{4}}{12}

\int \frac{1}{4 x} d x = \frac{1}{4} ln ∣ x ∣

= \frac{x ^{4}}{12} + \frac{1}{4} ln ∣ x ∣

Add a constant to the solution

= \frac{x ^{4}}{12} + \frac{1}{4} ln ∣ x ∣ + C

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What is the integral of (x^3)/3+1/(4x) ?
The integral of (x^3)/3+1/(4x) is (x^4)/(12)+1/4 ln|x|+C