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

The integral of 2xe^{xy}+x^2ye^{xy} is e^{yx}x^2+C

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integral of 2xe^{xy}+x^2ye^{xy}

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

e^{y x} x^{2} + C

Solution steps

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

Apply u-substitution

= \int \frac{2 e ^{u} u + e ^{u} u ^{2}}{y ^{2}} d u

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= \frac{1}{y ^{2}} \cdot \int 2 e^{u} u + e^{u} u^{2} d u

Factor

2 e^{u} u + e^{u} u^{2} : e^{u} u (2 + u)

= \frac{1}{y ^{2}} \cdot \int e^{u} u (2 + u) d u

Apply Integration By Parts

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

\int e^{u} u - e^{u} d u = e^{u} u - 2 e^{u}

= \frac{1}{y ^{2}} ((2 + u) (e^{u} u - e^{u}) - (e^{u} u - 2 e^{u}))

Substitute back

u = y x

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

Simplify

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

= e^{y x} x^{2}

Add a constant to the solution

= e^{y x} x^{2} + C

What is the integral of 2xe^{xy}+x^2ye^{xy} ?
The integral of 2xe^{xy}+x^2ye^{xy} is e^{yx}x^2+C