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

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

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

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

\frac{x ^{2}}{2} - 4 x + C

Solution steps

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

Expand

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

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

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

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

Add a constant to the solution

= \frac{x ^{2}}{2} - 4 x + C

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