What is the integral of-14x^{-4}y^6+16x^{-5}y^7 ?

The integral of-14x^{-4}y^6+16x^{-5}y^7 is -(2y^7)/(x^4)+(2y^8)/(x^5)+C

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integral of-14x^{-4}y^6+16x^{-5}y^7

\int - 14 x^{- 4} y^{6} + 16 x^{- 5} y^{7} d y

- \frac{2 y ^{7}}{x ^{4}} + \frac{2 y ^{8}}{x ^{5}} + C

Solution steps

\int - 14 x^{- 4} y^{6} + 16 x^{- 5} y^{7} 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 14 x^{- 4} y^{6} d y + \int 16 x^{- 5} y^{7} d y

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

\int 16 x^{- 5} y^{7} d y = \frac{2 y ^{8}}{x ^{5}}

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

Add a constant to the solution

= - \frac{2 y ^{7}}{x ^{4}} + \frac{2 y ^{8}}{x ^{5}} + C

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

y^{^{''}} - 14 y^{^{'}} + 51 y = 0

x y^{^{'}} + y = - 8 x y^{2}, y (1) = 1

\int \frac{7 x}{x ^{2} + 4} d x

(3 cos (x))^{^{'}}

What is the integral of-14x^{-4}y^6+16x^{-5}y^7 ?
The integral of-14x^{-4}y^6+16x^{-5}y^7 is -(2y^7)/(x^4)+(2y^8)/(x^5)+C