What is the integral of-8x^7y^{-5}+6x^6y^{-4} ?

The integral of-8x^7y^{-5}+6x^6y^{-4} is -(2x^6)/(y^3)+(2x^7)/(y^4)+C

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integral of-8x^7y^{-5}+6x^6y^{-4}

\int - 8 x^{7} y^{- 5} + 6 x^{6} y^{- 4} d y

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

Solution steps

\int - 8 x^{7} y^{- 5} + 6 x^{6} y^{- 4} 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 6 x^{6} y^{- 4} d y - \int 8 x^{7} y^{- 5} d y

\int 6 x^{6} y^{- 4} d y = - \frac{2 x ^{6}}{y ^{3}}

\int 8 x^{7} y^{- 5} d y = - \frac{2 x ^{7}}{y ^{4}}

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

Simplify

= - \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

\int (ln (x))^{10} \frac{1}{x} d x

y^{^{'''}} - y^{^{''}} + 8 y^{^{'}} - 8 y = 0

25 x e^{- x}

\frac{d ^{3}}{d x ^{3}} (2^{x})

\frac{\partial}{\partial x} (6 x y^{4} z^{3})

What is the integral of-8x^7y^{-5}+6x^6y^{-4} ?
The integral of-8x^7y^{-5}+6x^6y^{-4} is -(2x^6)/(y^3)+(2x^7)/(y^4)+C