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

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

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integral of (x^2-4)/x

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

\frac{x ^{2}}{2} - 4 ln ∣ x ∣ + C

Solution steps

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

Expand

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

= \int x - \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 d x - \int \frac{4}{x} d x

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

\int \frac{4}{x} d x = 4 ln ∣ x ∣

= \frac{x ^{2}}{2} - 4 ln ∣ x ∣

Add a constant to the solution

= \frac{x ^{2}}{2} - 4 ln ∣ x ∣ + C

\frac{d}{d x} (x^{2} 2 x - 3)

x^{2} y^{^{'}} + \frac{4}{3} x y = - 4 cos (3 x) y^{- \frac{1}{2}}

x \to 1 + lim (\frac{1}{x ^{2} - 1})

y^{^{''}} - 4 y = 0, y (0) = 0, y^{^{'}} (0) = 1

f (x) = \frac{4}{1 + x ^{2}}, (1, 2)

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