What is the integral of (3x-y)/9 ?

The integral of (3x-y)/9 is 1/9 (-(y^2)/2+3xy)+C

integral of (3x-y)/9

\int \frac{3 x - y}{9} d y

\frac{1}{9} (- \frac{y ^{2}}{2} + 3 x y) + C

Solution steps

\int \frac{3 x - y}{9} d y

Take the constant out:

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

= \frac{1}{9} \cdot \int - y + 3 x d y

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}{9} (- \int y d y + \int 3 x d y)

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

\int 3 x d y = 3 x y

= \frac{1}{9} (- \frac{y ^{2}}{2} + 3 x y)

Add a constant to the solution

= \frac{1}{9} (- \frac{y ^{2}}{2} + 3 x y) + C

y^{^{''}} - 14 y^{^{'}} + 49 y = 3 t^{- 1} e^{7 t}

s l o p e (- 3, 4), (2, - 9)

\int tan^{- \frac{3}{2}} (x) sec^{4} (x) d x

d er i v a t i v e g (x) = 4 x e^{x}

\frac{d y}{d x} = (1 + y^{2}) tan (x)

What is the integral of (3x-y)/9 ?
The integral of (3x-y)/9 is 1/9 (-(y^2)/2+3xy)+C