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受欢迎的微积分 >

integral of (x^2)/(sqrt((21+4x-x^2)))

解答

\int \frac{x ^{2}}{( 21 + 4 x - x ^{2} )} d x

解答

\frac{33}{2} arcsin (\frac{1}{5} (x - 2)) - \frac{25}{4} sin (2 arcsin (\frac{1}{5} (x - 2))) - 4 - x^{2} + 4 x + 21 + C

求解步骤

\int \frac{x ^{2}}{21 + 4 x - x ^{2}} d x

对

21 + 4 x - x^{2}

配方:

- (x - 2)^{2} + 25

= \int \frac{x ^{2}}{- ( x - 2 ) ^{2} + 25} d x

使用换元积分法

= \int \frac{( u + 2 ) ^{2}}{- u ^{2} + 25} d u

使用三角换元法

= \int (5 sin (v) + 2)^{2} d v

乘开

(5 sin (v) + 2)^{2} : 25 sin^{2} (v) + 20 sin (v) + 4

= \int 25 sin^{2} (v) + 20 sin (v) + 4 d v

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 25 sin^{2} (v) d v + \int 20 sin (v) d v + \int 4 d v

\int 25 sin^{2} (v) d v = \frac{25}{2} (v - \frac{1}{2} sin (2 v))

\int 20 sin (v) d v = - 20 cos (v)

\int 4 d v = 4 v

= \frac{25}{2} (v - \frac{1}{2} sin (2 v)) - 20 cos (v) + 4 v

代回

= \frac{25}{2} (arcsin (\frac{1}{5} (x - 2)) - \frac{1}{2} sin (2 arcsin (\frac{1}{5} (x - 2)))) - 20 cos (arcsin (\frac{1}{5} (x - 2))) + 4 arcsin (\frac{1}{5} (x - 2))

化简

\frac{25}{2} (arcsin (\frac{1}{5} (x - 2)) - \frac{1}{2} sin (2 arcsin (\frac{1}{5} (x - 2)))) - 20 cos (arcsin (\frac{1}{5} (x - 2))) + 4 arcsin (\frac{1}{5} (x - 2)) : \frac{33}{2} arcsin (\frac{1}{5} (x - 2)) - \frac{25}{4} sin (2 arcsin (\frac{1}{5} (x - 2))) - 4 - x^{2} + 4 x + 21

= \frac{33}{2} arcsin (\frac{1}{5} (x - 2)) - \frac{25}{4} sin (2 arcsin (\frac{1}{5} (x - 2))) - 4 - x^{2} + 4 x + 21

解答补常数

= \frac{33}{2} arcsin (\frac{1}{5} (x - 2)) - \frac{25}{4} sin (2 arcsin (\frac{1}{5} (x - 2))) - 4 - x^{2} + 4 x + 21 + C

作图

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