What is the integral of (4y^2-6y+x+2)/x ?

The integral of (4y^2-6y+x+2)/x is (2y)/x+y+(4y^3}{3x}-\frac{3y^2)/x+C

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integral of (4y^2-6y+x+2)/x

\int \frac{4 y ^{2} - 6 y + x + 2}{x}

\frac{2 y}{x} + y + \frac{4 y ^{3}}{3 x} - \frac{3 y ^{2}}{x} + C

Solution steps

\int \frac{4 y ^{2} - 6 y + x + 2}{x} d y

Take the constant out:

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

= \frac{1}{x} \cdot \int 2 + x + 4 y^{2} - 6 y 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}{x} (\int 2 d y + \int x d y + \int 4 y^{2} d y - \int 6 y d y)

\int 2 d y = 2 y

\int x d y = x y

\int 4 y^{2} d y = \frac{4 y ^{3}}{3}

\int 6 y d y = 3 y^{2}

= \frac{1}{x} (2 y + x y + \frac{4 y ^{3}}{3} - 3 y^{2})

Simplify

\frac{1}{x} (2 y + x y + \frac{4 y ^{3}}{3} - 3 y^{2}) : \frac{2 y}{x} + y + \frac{4 y ^{3}}{3 x} - \frac{3 y ^{2}}{x}

= \frac{2 y}{x} + y + \frac{4 y ^{3}}{3 x} - \frac{3 y ^{2}}{x}

Add a constant to the solution

= \frac{2 y}{x} + y + \frac{4 y ^{3}}{3 x} - \frac{3 y ^{2}}{x} + C

y^{^{''}} - 2 y^{^{'}} = 0

\frac{\partial}{\partial x} ((x + y) ln (x + y))

2 + 3 e^{- 3 t}

\int x^{5} e^{- x^{2}} d x

f (x) = \frac{2}{3} x^{3} + 4 x^{2} + 6 x - 5

What is the integral of (4y^2-6y+x+2)/x ?
The integral of (4y^2-6y+x+2)/x is (2y)/x+y+(4y^3}{3x}-\frac{3y^2)/x+C