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

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

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

\int (3 x^{3} - \frac{1}{4 x}) d x

\frac{3 x ^{4}}{4} - \frac{1}{4} ln ∣ x ∣ + C

Solution steps

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

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

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

= \frac{3 x ^{4}}{4} - \frac{1}{4} ln ∣ x ∣

Add a constant to the solution

= \frac{3 x ^{4}}{4} - \frac{1}{4} ln ∣ x ∣ + C

\int e^{x - 2} d x

n = 0 \sum \infty 9 (\frac{1 - 3}{4})^{n}

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

\int_{0}^{1} (- x + 2) - (x^{2} - 1) d x

\int x^{3} e^{x} d x

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