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

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

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

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

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

Solution steps

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

Simplify

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

= \int x + 2 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 2 d x

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

\int 2 d x = 2 x

= \frac{x ^{2}}{2} + 2 x

Add a constant to the solution

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

f (x) = (3 x^{2} + 2)^{2} (x^{2} - 5 x)^{3}

\int x^{2} ln (x^{3}) d x

x \to - 1 - lim (x^{3} + 2 x^{2} + 1)

\int x^{4} ln (5 x) d x

(ln (x^{3}))^{^{'}}

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