integral of xsqrt(x+9)

解答

\int x x + 9 d x

\frac{2}{5} (x + 9)^{\frac{5}{2}} - 6 (x + 9)^{\frac{3}{2}} + C

求解步骤

\int x x + 9 d x

使用换元积分法

= \int (u - 9) u d u

乘开

(u - 9) u : u^{\frac{3}{2}} - 9 u

= \int u^{\frac{3}{2}} - 9 u d u

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 u^{\frac{3}{2}} d u - \int 9 u d u

\int u^{\frac{3}{2}} d u = \frac{2}{5} u^{\frac{5}{2}}

\int 9 u d u = 6 u^{\frac{3}{2}}

= \frac{2}{5} u^{\frac{5}{2}} - 6 u^{\frac{3}{2}}

u = x + 9

代回

= \frac{2}{5} (x + 9)^{\frac{5}{2}} - 6 (x + 9)^{\frac{3}{2}}

解答补常数

= \frac{2}{5} (x + 9)^{\frac{5}{2}} - 6 (x + 9)^{\frac{3}{2}} + C

x \to - 1 lim (x^{2} - 5 x + 6)

\frac{\partial}{\partial y} (3 x^{2} - 3 y)

\int \frac{( x ^{2} - 9 )}{x ^{3}} d x

\int \frac{tan ( x ) + 1}{tan ( x ) - 1} d x

\frac{d y}{d x} y = \frac{9 x - 6}{54 x ^{2}}