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

The integral of (x^3-3x^2+4x-1)/(3x) is (x^3)/9-(x^2)/2+4/3 x-1/3 ln|x|+C

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integral of (x^3-3x^2+4x-1)/(3x)

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

\frac{x ^{3}}{9} - \frac{x ^{2}}{2} + \frac{4}{3} x - \frac{1}{3} ln ∣ x ∣ + C

Solution steps

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

Take the constant out:

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

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

Expand

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

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

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

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

\int 4 d x = 4 x

\int \frac{1}{x} d x = ln ∣ x ∣

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

Simplify

\frac{1}{3} (\frac{x ^{3}}{3} - \frac{3 x ^{2}}{2} + 4 x - ln ∣ x ∣) : \frac{x ^{3}}{9} - \frac{x ^{2}}{2} + \frac{4}{3} x - \frac{1}{3} ln ∣ x ∣

= \frac{x ^{3}}{9} - \frac{x ^{2}}{2} + \frac{4}{3} x - \frac{1}{3} ln ∣ x ∣

Add a constant to the solution

= \frac{x ^{3}}{9} - \frac{x ^{2}}{2} + \frac{4}{3} x - \frac{1}{3} ln ∣ x ∣ + C

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What is the integral of (x^3-3x^2+4x-1)/(3x) ?
The integral of (x^3-3x^2+4x-1)/(3x) is (x^3)/9-(x^2)/2+4/3 x-1/3 ln|x|+C