What is the integral of 2xe^y+2xy^{-1}+y^{-3} ?

The integral of 2xe^y+2xy^{-1}+y^{-3} is x/(y^3)+(x^2)/y+e^yx^2+C

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integral of 2xe^y+2xy^{-1}+y^{-3}

\int 2 x e^{y} + 2 x y^{- 1} + y^{- 3}

\frac{x}{y ^{3}} + \frac{x ^{2}}{y} + e^{y} x^{2} + C

Solution steps

\int 2 x e^{y} + 2 x y^{- 1} + y^{- 3} d x

Simplify

y^{- 3} + 2 y^{- 1} x + 2 e^{y} x : \frac{1}{y ^{3}} + \frac{2 x}{y} + 2 e^{y} x

= \int \frac{1}{y ^{3}} + \frac{2 x}{y} + 2 e^{y} x 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 \frac{1}{y ^{3}} d x + \int \frac{2 x}{y} d x + \int 2 e^{y} x d x

\int \frac{1}{y ^{3}} d x = \frac{x}{y ^{3}}

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

\int 2 e^{y} x d x = e^{y} x^{2}

= \frac{x}{y ^{3}} + \frac{x ^{2}}{y} + e^{y} x^{2}

Add a constant to the solution

= \frac{x}{y ^{3}} + \frac{x ^{2}}{y} + e^{y} x^{2} + C

\frac{- 2 s + 6}{s ^{2} + 4}

x \to 2 lim (ln (\frac{x}{x - 2}))

y^{'} = 3 y \cdot (y - 2)

x \to π lim (\frac{sin ( x )}{1 - \frac{x ^{2}}{π ^{2}}})

x \to \infty lim (9 x^{2} e^{- x})

What is the integral of 2xe^y+2xy^{-1}+y^{-3} ?
The integral of 2xe^y+2xy^{-1}+y^{-3} is x/(y^3)+(x^2)/y+e^yx^2+C