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

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

\int 12 y^{- 3} x^{5} d x = \frac{2 x ^{6}}{y ^{3}}

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

= - \frac{2 x ^{6}}{y ^{3}} + \frac{2 x ^{7}}{y ^{4}}

Add a constant to the solution

= - \frac{2 x ^{6}}{y ^{3}} + \frac{2 x ^{7}}{y ^{4}} + C

x \to 3 lim (12 - x - 3)

x \to \infty lim ((1 + (\frac{2}{x}))^{(\frac{x}{4})})

\int_{0}^{3} x^{3} - 5 x d x

\frac{d y}{d x} = \frac{1}{7} y cos^{2} (y)

(y sin (x) + x y cos (x)) d x + (x sin (x) + 1) d y = 0

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