What is the integral of xe^{xy}+2xy+1/x ?

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

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integral of xe^{xy}+2xy+1/x

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

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

Solution steps

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

\int \frac{1}{x} d x = ln ∣ x ∣

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

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

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

Add a constant to the solution

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

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

\int_{- 3}^{5} (1 + 4 x) d x

\int sin (x) sin (x) d x

y = ln (x^{2}), at x = 4

\frac{\partial}{\partial y} (\frac{1}{x y ^{2} - x ^{2} y})

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