What is the integral of ((x-2))/1 ?

The integral of ((x-2))/1 is (x^2)/2-2x+C

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integral of ((x-2))/1

\int \frac{( x - 2 )}{1} d x

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

Solution steps

\int \frac{x - 2}{1} d x

Simplify

= \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

\frac{d}{d x} (9 arctan (x + 1 + x^{2}))

x \to 4 lim (\frac{4 x}{x - 4})

\int \frac{sin ( x )}{x ^{2}} d x

\int \frac{x + 1}{2 x ^{2} + 4 x + 5} d x

y^{^{'}} = y sin (2 x), y (0) = 1.6