integral of t^{-1/2}(t^2-t+2)

解答

\int t^{- \frac{1}{2}} (t^{2} - t + 2) d t

\frac{2}{5} t^{\frac{5}{2}} - \frac{2}{3} t^{\frac{3}{2}} + 4 t^{\frac{1}{2}} + C

求解步骤

\int t^{- \frac{1}{2}} (t^{2} - t + 2) d t

使用分布积分法

= (t^{2} - t + 2) \cdot 2 t^{\frac{1}{2}} - \int (2 t - 1) \cdot 2 t^{\frac{1}{2}} d t

\int (2 t - 1) \cdot 2 t^{\frac{1}{2}} d t = 2 (\frac{4}{5} t^{\frac{5}{2}} - \frac{2}{3} t^{\frac{3}{2}})

= (t^{2} - t + 2) \cdot 2 t^{\frac{1}{2}} - 2 (\frac{4}{5} t^{\frac{5}{2}} - \frac{2}{3} t^{\frac{3}{2}})

化简

(t^{2} - t + 2) \cdot 2 t^{\frac{1}{2}} - 2 (\frac{4}{5} t^{\frac{5}{2}} - \frac{2}{3} t^{\frac{3}{2}}) : \frac{2}{5} t^{\frac{5}{2}} - \frac{2}{3} t^{\frac{3}{2}} + 4 t^{\frac{1}{2}}

= \frac{2}{5} t^{\frac{5}{2}} - \frac{2}{3} t^{\frac{3}{2}} + 4 t^{\frac{1}{2}}

解答补常数

= \frac{2}{5} t^{\frac{5}{2}} - \frac{2}{3} t^{\frac{3}{2}} + 4 t^{\frac{1}{2}} + C

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