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

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

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

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

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

Solution steps

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

Expand

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

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

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

\int 3 d x = 3 x

\int \frac{4}{x} d x = 4 ln ∣ x ∣

= \frac{x ^{3}}{3} - 3 x + 4 ln ∣ x ∣

Add a constant to the solution

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

\int_{- 2}^{2} y^{2} - 4 d y

f (x) = 4 cos (2 x)

\frac{d}{d x} (\frac{3}{x - 5})

y = x^{2} + 8

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