What is the integral of (x+1)sin(2x-3) ?

The integral of (x+1)sin(2x-3) is 1/4 (-cos(2x-3)(2x+2)+sin(2x-3))+C

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integral of (x+1)sin(2x-3)

\int (x + 1) sin (2 x - 3) d x

\frac{1}{4} (- cos (2 x - 3) (2 x + 2) + sin (2 x - 3)) + C

Solution steps

\int (x + 1) sin (2 x - 3) d x

Apply u-substitution

: \int \frac{sin ( u ) ( u + 5 )}{4} d u

= \int \frac{sin ( u ) ( u + 5 )}{4} d u

Take the constant out:

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

= \frac{1}{4} \cdot \int sin (u) (u + 5) d u

Apply Integration By Parts

: - cos (u) (u + 5) - \int - cos (u) d u

= \frac{1}{4} (- cos (u) (u + 5) - \int - cos (u) d u)

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

= \frac{1}{4} (- cos (u) (u + 5) - (- sin (u)))

Substitute back

u = 2 x - 3

= \frac{1}{4} (- cos (2 x - 3) (2 x - 3 + 5) - (- sin (2 x - 3)))

Simplify

\frac{1}{4} (- cos (2 x - 3) (2 x - 3 + 5) - (- sin (2 x - 3))) : \frac{1}{4} (- cos (2 x - 3) (2 x + 2) + sin (2 x - 3))

= \frac{1}{4} (- cos (2 x - 3) (2 x + 2) + sin (2 x - 3))

Add a constant to the solution

= \frac{1}{4} (- cos (2 x - 3) (2 x + 2) + sin (2 x - 3)) + C

y^{^{''}} - 20 y^{^{'}} + 181 y = 0

\int (x^{2} - 5)^{2} d x

\int \frac{x}{x} d x

y = 11 x^{- 2} + 7 x^{3} - 5 x

x \to 2 lim (4)

What is the integral of (x+1)sin(2x-3) ?
The integral of (x+1)sin(2x-3) is 1/4 (-cos(2x-3)(2x+2)+sin(2x-3))+C