What is the integral of xycos(xy)+sin(xy) ?

The integral of xycos(xy)+sin(xy) is xsin(yx)+C

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integral of xycos(xy)+sin(xy)

\int x y cos (x y) + sin (x y) d x

x sin (y x) + C

Solution steps

\int x y cos (x y) + sin (x y) d x

Apply u-substitution

: \int \frac{sin ( u ) + u cos ( u )}{y} d u

= \int \frac{sin ( u ) + u cos ( u )}{y} d u

Take the constant out:

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

= \frac{1}{y} \cdot \int sin (u) + u cos (u) d u

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= \frac{1}{y} (\int sin (u) d u + \int u cos (u) d u)

\int sin (u) d u = - cos (u)

\int u cos (u) d u = u sin (u) + cos (u)

= \frac{1}{y} (- cos (u) + u sin (u) + cos (u))

Substitute back

u = y x

= \frac{1}{y} (- cos (y x) + y x sin (y x) + cos (y x))

Simplify

\frac{1}{y} (- cos (y x) + y x sin (y x) + cos (y x)) : x sin (y x)

= x sin (y x)

Add a constant to the solution

= x sin (y x) + C

\frac{\partial}{\partial x} (21 - 3 x^{2} - 4 y^{2})

\frac{d}{d x} (7^{- 4 x + 1 + 2 x^{- 1}})

\frac{d}{d x} (x (- \frac{1}{10} x + 610))

x \to 1 lim (2^{0})

\int_{2}^{3} 1 d x

What is the integral of xycos(xy)+sin(xy) ?
The integral of xycos(xy)+sin(xy) is xsin(yx)+C