What is the integral of (6x^2y-3xsqrt(y)) ?

The integral of (6x^2y-3xsqrt(y)) is -2xy^{3/2}+3x^2y^2+C

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integral of (6x^2y-3xsqrt(y))

\int (6 x^{2} y - 3 x y) d y

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

Solution steps

\int 6 x^{2} y - 3 x 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

= - \int 3 y x d y + \int 6 x^{2} y d y

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

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

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

Add a constant to the solution

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

\int 3 sin (2 θ) d θ

3 x^{4} + 9 x^{3} + 11

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

y^{^{''}} - 6 y^{^{'}} + 13 = 0

(- 4, 2), (1, - 4)

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