题目 Problem Find ∫(10x4−32x3−7) dx\begin{align*} \int\left(10x^4-\frac{3}{2x^3}-7\right)\,dx \end{align*}∫(10x4−2x33−7)dx giving each term in simplest form. (3) 解答 解法一 思路 展开 先把分式写成指数形式： 32x3=32x−3.\begin{align*} \frac{3}{2x^3}=\frac32x^{-3}. \end{align*}2x33=23x−3. 然后逐项积分，最后不要忘记 +c+c+c。 答题过程 展开 ∫(10x4−32x3−7) dx= ∫(10x4−32x−3−7) dx= 10⋅x55−32⋅x−2−2−7x+c= 2x5+34x−2−7x+c.\begin{align*} \int\left(10x^4-\frac{3}{2x^3}-7\right)\,dx =&\,\int\left(10x^4-\frac32x^{-3}-7\right)\,dx\\ =&\,10\cdot\frac{x^5}{5} -\frac32\cdot\frac{x^{-2}}{-2} -7x+c\\ =&\,2x^5+\frac34x^{-2}-7x+c. \end{align*}∫(10x4−2x33−7)dx===∫(10x4−23x−3−7)dx10⋅5x5−23⋅−2x−2−7x+c2x5+43x−2−7x+c. So ∫(10x4−32x3−7) dx=2x5+34x2−7x+c.\begin{align*} \int\left(10x^4-\frac{3}{2x^3}-7\right)\,dx =2x^5+\frac{3}{4x^2}-7x+c. \end{align*}∫(10x4−2x33−7)dx=2x5+4x23−7x+c.
