What is the integral of (x^4+3x^2-2)/(4x) ?

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

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integral of (x^4+3x^2-2)/(4x)

\int \frac{x ^{4} + 3 x ^{2} - 2}{4 x} d x

\frac{1}{4} (\frac{x ^{4}}{4} + \frac{3 x ^{2}}{2} - 2 ln ∣ x ∣) + C

Solution steps

\int \frac{x ^{4} + 3 x ^{2} - 2}{4 x} d x

Take the constant out:

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

= \frac{1}{4} \cdot \int \frac{x ^{4} + 3 x ^{2} - 2}{x} d x

Expand

\frac{x ^{4} + 3 x ^{2} - 2}{x} : x^{3} + 3 x - \frac{2}{x}

= \frac{1}{4} \cdot \int x^{3} + 3 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}{4} (\int x^{3} d x + \int 3 x d x - \int \frac{2}{x} d x)

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

\int 3 x d x = \frac{3 x ^{2}}{2}

\int \frac{2}{x} d x = 2 ln ∣ x ∣

= \frac{1}{4} (\frac{x ^{4}}{4} + \frac{3 x ^{2}}{2} - 2 ln ∣ x ∣)

Add a constant to the solution

= \frac{1}{4} (\frac{x ^{4}}{4} + \frac{3 x ^{2}}{2} - 2 ln ∣ x ∣) + C

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