题目 Problem Find, in simplest form, ∫(8x33−12x−5) dx.\begin{align*} \int\left(\frac{8x^3}{3}-\frac{1}{2\sqrt{x}}-5\right)\,dx. \end{align*}∫(38x3−2x1−5)dx. (4) 解答 解法一 思路 展开 先把根号写成指数形式，再逐项积分。最后不要忘记积分常数。 答题过程 展开 ∫(8x33−12x−5) dx= ∫(83x3−12x−1/2−5) dx= 83⋅x44−12⋅x1/21/2−5x+c= 23x4−x1/2−5x+c= 23x4−x−5x+c.\begin{align*} \int\left(\frac{8x^3}{3}-\frac{1}{2\sqrt{x}}-5\right)\,dx =&\,\int\left(\frac83x^3-\frac12x^{-1/2}-5\right)\,dx\\ =&\,\frac83\cdot\frac{x^4}{4} -\frac12\cdot\frac{x^{1/2}}{1/2} -5x+c\\ =&\,\frac23x^4-x^{1/2}-5x+c\\ =&\,\frac23x^4-\sqrt{x}-5x+c. \end{align*}∫(38x3−2x1−5)dx====∫(38x3−21x−1/2−5)dx38⋅4x4−21⋅1/2x1/2−5x+c32x4−x1/2−5x+c32x4−x−5x+c.
