What is the integral of (2x-3)/6 ?

The integral of (2x-3)/6 is 1/6 (x^2-3x)+C

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integral of (2x-3)/6

\int \frac{2 x - 3}{6} d x

\frac{1}{6} (x^{2} - 3 x) + C

Solution steps

\int \frac{2 x - 3}{6} d x

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= \frac{1}{6} \cdot \int 2 x - 3 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

= \frac{1}{6} (\int 2 x d x - \int 3 d x)

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

\int 3 d x = 3 x

= \frac{1}{6} (x^{2} - 3 x)

Add a constant to the solution

= \frac{1}{6} (x^{2} - 3 x) + C

ln (x^{2})

\frac{1}{( s + 3 )}

\int \frac{1}{( 2 x - 3 ) \cdot ( 3 x ^{2} + x )} d x

\frac{d}{d x} (\frac{x ^{2}}{x ^{4}})

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