integral of x/((x+2)^{1/4)}

解答

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

\frac{4}{7} (x + 2)^{\frac{7}{4}} - \frac{8}{3} (x + 2)^{\frac{3}{4}} + C

求解步骤

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

使用换元积分法

= \int \frac{u - 2}{u ^{\frac{1}{4}}} d u

乘开

\frac{u - 2}{u ^{\frac{1}{4}}} : u^{\frac{3}{4}} - \frac{2}{u ^{\frac{1}{4}}}

= \int u^{\frac{3}{4}} - \frac{2}{u ^{\frac{1}{4}}} 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}{4}} d u - \int \frac{2}{u ^{\frac{1}{4}}} d u

\int u^{\frac{3}{4}} d u = \frac{4}{7} u^{\frac{7}{4}}

\int \frac{2}{u ^{\frac{1}{4}}} d u = \frac{8}{3} u^{\frac{3}{4}}

= \frac{4}{7} u^{\frac{7}{4}} - \frac{8}{3} u^{\frac{3}{4}}

u = x + 2

代回

= \frac{4}{7} (x + 2)^{\frac{7}{4}} - \frac{8}{3} (x + 2)^{\frac{3}{4}}

解答补常数

= \frac{4}{7} (x + 2)^{\frac{7}{4}} - \frac{8}{3} (x + 2)^{\frac{3}{4}} + C

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