What is the integral of (x^2+x+1)*sin(x) ?

The integral of (x^2+x+1)*sin(x) is -x^2cos(x)-xcos(x)+2xsin(x)+cos(x)+sin(x)+C

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integral of (x^2+x+1)*sin(x)

\int (x^{2} + x + 1) \cdot sin (x) d x

- x^{2} cos (x) - x cos (x) + 2 x sin (x) + cos (x) + sin (x) + C

Solution steps

\int (x^{2} + x + 1) sin (x) d x

Apply Integration By Parts

: - cos (x) (x^{2} + x + 1) - \int - cos (x) (2 x + 1) d x

= - cos (x) (x^{2} + x + 1) - \int - cos (x) (2 x + 1) d x

\int - cos (x) (2 x + 1) d x = - 2 x sin (x) - sin (x) - 2 cos (x)

= - cos (x) (x^{2} + x + 1) - (- 2 x sin (x) - sin (x) - 2 cos (x))

Simplify

- cos (x) (x^{2} + x + 1) - (- 2 x sin (x) - sin (x) - 2 cos (x)) : - x^{2} cos (x) - x cos (x) + 2 x sin (x) + cos (x) + sin (x)

= - x^{2} cos (x) - x cos (x) + 2 x sin (x) + cos (x) + sin (x)

Add a constant to the solution

= - x^{2} cos (x) - x cos (x) + 2 x sin (x) + cos (x) + sin (x) + C

\int \frac{arctan ( 4 x )}{1 + 16 x ^{2}} d x

y = 5 x - 2 x^{2}, (- 3, - 33)

\int 1 + 2 cos (2 θ) + \frac{1}{2} (1 + cos (4 θ)) d θ

(12 x^{2} + y - 1) d x - (10 y - x) d y = 0, y (1) = 0

\frac{9}{4 ( 3 x + 2 ) ^{\frac{3}{2}}}

What is the integral of (x^2+x+1)*sin(x) ?
The integral of (x^2+x+1)*sin(x) is -x^2cos(x)-xcos(x)+2xsin(x)+cos(x)+sin(x)+C