integral of (x^2+7x-3)sin(2x)

解答

\int (x^{2} + 7 x - 3) sin (2 x) d x

- \frac{1}{2} x^{2} cos (2 x) + \frac{1}{2} x sin (2 x) - \frac{7}{2} x cos (2 x) + \frac{7}{4} cos (2 x) + \frac{7}{4} sin (2 x) + C

求解步骤

\int (x^{2} + 7 x - 3) sin (2 x) d x

乘开

(x^{2} + 7 x - 3) sin (2 x) : x^{2} sin (2 x) + 7 x sin (2 x) - 3 sin (2 x)

= \int x^{2} sin (2 x) + 7 x sin (2 x) - 3 sin (2 x) d x

使用积分加法定则:

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

= \int x^{2} sin (2 x) d x + \int 7 x sin (2 x) d x - \int 3 sin (2 x) d x

\int x^{2} sin (2 x) d x = - \frac{1}{2} x^{2} cos (2 x) + \frac{1}{4} (2 x sin (2 x) + cos (2 x))

\int 7 x sin (2 x) d x = 7 (- \frac{1}{2} x cos (2 x) + \frac{1}{4} sin (2 x))

\int 3 sin (2 x) d x = - \frac{3}{2} cos (2 x)

= - \frac{1}{2} x^{2} cos (2 x) + \frac{1}{4} (2 x sin (2 x) + cos (2 x)) + 7 (- \frac{1}{2} x cos (2 x) + \frac{1}{4} sin (2 x)) - (- \frac{3}{2} cos (2 x))

化简

- \frac{1}{2} x^{2} cos (2 x) + \frac{1}{4} (2 x sin (2 x) + cos (2 x)) + 7 (- \frac{1}{2} x cos (2 x) + \frac{1}{4} sin (2 x)) - (- \frac{3}{2} cos (2 x)) : - \frac{1}{2} x^{2} cos (2 x) + \frac{1}{2} x sin (2 x) - \frac{7}{2} x cos (2 x) + \frac{7}{4} cos (2 x) + \frac{7}{4} sin (2 x)

= - \frac{1}{2} x^{2} cos (2 x) + \frac{1}{2} x sin (2 x) - \frac{7}{2} x cos (2 x) + \frac{7}{4} cos (2 x) + \frac{7}{4} sin (2 x)

解答补常数

= - \frac{1}{2} x^{2} cos (2 x) + \frac{1}{2} x sin (2 x) - \frac{7}{2} x cos (2 x) + \frac{7}{4} cos (2 x) + \frac{7}{4} sin (2 x) + C

\int \frac{3}{x ( x ^{2} + 1 )} d x

\frac{d}{d x} (∣ (- 3)^{x} ∣)

y^{^{'}} x^{2} + y x^{2} = x y^{5}

d er i v a t i v e f (x) = \frac{1}{2}

d er i v a t i v e \frac{2 x ^{2} - 1}{x + 1}