What is the integral of xsin(3+3x) ?

The integral of xsin(3+3x) is 1/9 (-3xcos(3+3x)+sin(3+3x))+C

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integral of xsin(3+3x)

\int x sin (3 + 3 x) d x

\frac{1}{9} (- 3 x cos (3 + 3 x) + sin (3 + 3 x)) + C

Solution steps

\int x sin (3 + 3 x) d x

Apply u-substitution

: \int \frac{sin ( u ) ( u - 3 )}{9} d u

= \int \frac{sin ( u ) ( u - 3 )}{9} d u

Take the constant out:

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

= \frac{1}{9} \cdot \int sin (u) (u - 3) d u

Apply Integration By Parts

: - cos (u) (u - 3) - \int - cos (u) d u

= \frac{1}{9} (- cos (u) (u - 3) - \int - cos (u) d u)

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

= \frac{1}{9} (- cos (u) (u - 3) - (- sin (u)))

Substitute back

u = 3 + 3 x

= \frac{1}{9} (- cos (3 + 3 x) (3 + 3 x - 3) - (- sin (3 + 3 x)))

Simplify

\frac{1}{9} (- cos (3 + 3 x) (3 + 3 x - 3) - (- sin (3 + 3 x))) : \frac{1}{9} (- 3 x cos (3 + 3 x) + sin (3 + 3 x))

= \frac{1}{9} (- 3 x cos (3 + 3 x) + sin (3 + 3 x))

Add a constant to the solution

= \frac{1}{9} (- 3 x cos (3 + 3 x) + sin (3 + 3 x)) + C

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

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

\frac{d}{d x} (ln (10 x^{2} - 8))

y^{^{'}} = e^{2 t + y - 1} - 2

\frac{\partial}{\partial x} (e^{x y} + ln (\frac{x}{y}))

What is the integral of xsin(3+3x) ?
The integral of xsin(3+3x) is 1/9 (-3xcos(3+3x)+sin(3+3x))+C