What is the integral of (e^{yx}+ye^{yx}x+2y-1/(x^2)) ?

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

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integral of (e^{yx}+ye^{yx}x+2y-1/(x^2))

\int (e^{y x} + y e^{y x} x + 2 y - \frac{1}{x ^{2}}) d y

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

Solution steps

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

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

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

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

\int x e^{x y} y d y = e^{x y} y - \frac{e ^{x y}}{x}

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

Add a constant to the solution

= - \frac{y}{x ^{2}} + \frac{1}{x} e^{x y} + y^{2} + e^{x y} y - \frac{e ^{x y}}{x} + C

\frac{d}{d x} (x (1 - x)^{6})

\int_{0}^{2} (x^{2} - 10 x) d x

\int ln (2 x) (x) d x

4 x^{4} - 4 x^{2}, 6 x^{2}

x \to 0 + lim (\frac{1}{x} sin (\frac{1}{x}))

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