What is the integral of 14x^6y^{-4}-12x^{5y^{-3}} ?

The integral of 14x^6y^{-4}-12x^{5y^{-3}} is -(12y^3x^{\frac{5+y^3)/(y^3)}}{5+y^3}+(2x^7)/(y^4)+C

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integral of 14x^6y^{-4}-12x^{5y^{-3}}

\int 14 x^{6} y^{- 4} - 12 x^{5 y^{- 3}} d x

- \frac{12 y ^{3} x ^{\frac{5 + y ^{3}}{y ^{3}}}}{5 + y ^{3}} + \frac{2 x ^{7}}{y ^{4}} + C

Solution steps

\int 14 x^{6} y^{- 4} - 12 x^{5 y^{- 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

= - \int 12 x^{5 y^{- 3}} d x + \int 14 y^{- 4} x^{6} d x

\int 12 x^{5 y^{- 3}} d x = \frac{12 y ^{3} x ^{\frac{5 + y ^{3}}{y ^{3}}}}{5 + y ^{3}}

\int 14 y^{- 4} x^{6} d x = \frac{2 x ^{7}}{y ^{4}}

= - \frac{12 y ^{3} x ^{\frac{5 + y ^{3}}{y ^{3}}}}{5 + y ^{3}} + \frac{2 x ^{7}}{y ^{4}}

Add a constant to the solution

= - \frac{12 y ^{3} x ^{\frac{5 + y ^{3}}{y ^{3}}}}{5 + y ^{3}} + \frac{2 x ^{7}}{y ^{4}} + C

\frac{\partial}{\partial x} (cos (2 x - y))

\int x ln (x^{6}) d x

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

x = - 5, x = 3, y = 10 x, y = x^{2} - 11

\frac{ln ( x )}{2} + 2

What is the integral of 14x^6y^{-4}-12x^{5y^{-3}} ?
The integral of 14x^6y^{-4}-12x^{5y^{-3}} is -(12y^3x^{\frac{5+y^3)/(y^3)}}{5+y^3}+(2x^7)/(y^4)+C