integral of x^2sqrt((x+1))

解答

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

\frac{2}{7} (x + 1)^{\frac{7}{2}} - \frac{4}{5} (x + 1)^{\frac{5}{2}} + \frac{2}{3} (x + 1)^{\frac{3}{2}} + C

求解步骤

\int x^{2} x + 1 d x

使用换元积分法

= \int (u - 1)^{2} u d u

乘开

(u - 1)^{2} u : u^{\frac{5}{2}} - 2 u^{\frac{3}{2}} + u

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

\int u^{\frac{5}{2}} d u = \frac{2}{7} u^{\frac{7}{2}}

\int 2 u^{\frac{3}{2}} d u = \frac{4}{5} u^{\frac{5}{2}}

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

= \frac{2}{7} u^{\frac{7}{2}} - \frac{4}{5} u^{\frac{5}{2}} + \frac{2}{3} u^{\frac{3}{2}}

u = x + 1

代回

= \frac{2}{7} (x + 1)^{\frac{7}{2}} - \frac{4}{5} (x + 1)^{\frac{5}{2}} + \frac{2}{3} (x + 1)^{\frac{3}{2}}

解答补常数

= \frac{2}{7} (x + 1)^{\frac{7}{2}} - \frac{4}{5} (x + 1)^{\frac{5}{2}} + \frac{2}{3} (x + 1)^{\frac{3}{2}} + C

\int \frac{1}{x ( 1 0 ^{- 1} - 1 0 ^{- 7} x )} d x

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

\int \frac{5}{( 6 + x ) ^{4}} d x

\int e^{- 3 r} d r

\int \frac{x ^{3} + x ^{2} + 5 x + 4}{( x ^{2} + 4 ) ^{2}} d x