What is the integral of 5x^4y^3-x^2y^6+6x^5-4y ?

The integral of 5x^4y^3-x^2y^6+6x^5-4y is -4yx+x^6-(y^6x^3)/3+y^3x^5+C

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integral of 5x^4y^3-x^2y^6+6x^5-4y

\int 5 x^{4} y^{3} - x^{2} y^{6} + 6 x^{5} - 4 y d x

- 4 y x + x^{6} - \frac{y ^{6} x ^{3}}{3} + y^{3} x^{5} + C

Solution steps

\int 5 x^{4} y^{3} - x^{2} y^{6} + 6 x^{5} - 4 y 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 4 y d x + \int 6 x^{5} d x - \int y^{6} x^{2} d x + \int 5 y^{3} x^{4} d x

\int 4 y d x = 4 y x

\int 6 x^{5} d x = x^{6}

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

\int 5 y^{3} x^{4} d x = y^{3} x^{5}

= - 4 y x + x^{6} - \frac{y ^{6} x ^{3}}{3} + y^{3} x^{5}

Add a constant to the solution

= - 4 y x + x^{6} - \frac{y ^{6} x ^{3}}{3} + y^{3} x^{5} + C

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

x \to 0 + lim (x ln (x))

\frac{d}{d x} (sin (6 x) cos (4 x))

x^{4} + y^{3} x^{5} + 5 y^{^{'}} = 0

3 x^{2} y^{2} d x + 2 x^{3} y d y = 0

What is the integral of 5x^4y^3-x^2y^6+6x^5-4y ?
The integral of 5x^4y^3-x^2y^6+6x^5-4y is -4yx+x^6-(y^6x^3)/3+y^3x^5+C