微積分學/不定積分/練習答案

1. ${\displaystyle \int {\frac {3x}{2}}dx}$
解答：欲找到函數${\displaystyle F}$，使
${\displaystyle F'(x)={\frac {3x}{2}}}$
由於
${\displaystyle {\frac {d}{dx}}x^{2}=2x}$
因此須找到常數${\displaystyle a}$，使
${\displaystyle {\frac {d}{dx}}ax^{2}=2ax={\frac {3x}{2}}}$
解出${\displaystyle a}$，得
${\displaystyle 2ax={\frac {3x}{2}}\implies a={\frac {3}{4}}}$

${\displaystyle \int {\frac {3x}{2}}={\frac {3}{4}}x^{2}+C}$
求導驗證：
${\displaystyle {\frac {d}{dx}}\left({\frac {3}{4}}x^{2}+C\right)={\frac {3}{4}}(2x)={\frac {3x}{2}}}$

2. ${\displaystyle f(x)=2x^{4}}$的原函數
解答：由於
${\displaystyle {\frac {d}{dx}}x^{5}=5x^{4}}$
須找到常數${\displaystyle a}$，使
${\displaystyle {\frac {d}{dx}}ax^{5}=5ax^{4}=2x^{4}}$
解出${\displaystyle a}$，得
${\displaystyle 5ax^{4}=2x^{4}\implies a={\frac {2}{5}}}$
故原函數為
${\displaystyle \mathbf {{\frac {2}{5}}x^{5}+C} }$
求導驗證：
${\displaystyle {\frac {d}{dx}}\int 2x^{4}dx={\frac {d}{dx}}({\frac {2}{5}}x^{5}+C)={\frac {2}{5}}(5x^{4})=2x^{4}}$

3. ${\displaystyle \int (7x^{2}+3\cos x-e^{x})dx}$
解答：
{\displaystyle {\begin{aligned}\int (7x^{2}+3\cos x-e^{x})dx&=7\int x^{2}dx+3\int \cos xdx-\int e^{x}dx\\&=7({\frac {x^{3}}{3}})+3\sin x-e^{x}+C\\&=\mathbf {{\frac {7}{3}}x^{3}+3\sin x-e^{x}+C} \end{aligned}}}

4. ${\displaystyle \int ({\frac {2}{5x}}+\sin x)dx}$
解答：
{\displaystyle {\begin{aligned}\int ({\frac {2}{5x}}+\sin x)dx&={\frac {2}{5}}\int {\frac {dx}{x}}+\int \sin xdx\\&=\mathbf {{\frac {2}{5}}\ln |x|-\cos x+C} \end{aligned}}}

5. ${\displaystyle \int x\sin 2x^{2}dx}$
解答：令${\displaystyle u=2x^{2}}$${\displaystyle du=4xdx}$${\displaystyle dx={\frac {du}{4x}}}$
{\displaystyle {\begin{aligned}\int x\sin 2x^{2}dx&=\int x\sin u{\frac {du}{4x}}\\&={\frac {1}{4}}\int \sin udu\\&=-{\frac {\cos u}{4}}+C\\&=-\mathbf {{\frac {\cos 2x^{2}}{4}}+C} \end{aligned}}}
6. ${\displaystyle \int -3e^{\sin x}\cos xdx}$
解答：令${\displaystyle u=\sin x}$${\displaystyle du=\cos xdx}$，則${\displaystyle dx={\frac {du}{\cos x}}}$
{\displaystyle {\begin{aligned}\int -3e^{\sin x}\cos xdx&=-3\int e^{u}\cos x{\frac {du}{\cos x}}\\&=-3\int e^{u}du\\&=-3e^{u}+C\\&=\mathbf {-3e^{\sin x}+C} \end{aligned}}}
7. ${\displaystyle \int {\frac {2x-5}{x^{3}}}dx}$
解答：令${\displaystyle u=2(x-5)}$${\displaystyle v=\int {\frac {dx}{x^{3}}}=-{\frac {1}{2x^{2}}}}$
{\displaystyle {\begin{aligned}\int {\frac {2x-5}{x^{3}}}dx&=\int udv\\&=uv-\int vdu\\&=(2x-5)(-{\frac {1}{2x^{2}}})-\int (-{\frac {1}{2x^{2}}})2dx\\&={\frac {5-2x}{2x^{2}}}+\int {\frac {dx}{x^{2}}}\\&={\frac {5-2x}{2x^{2}}}-{\frac {1}{x}}\\&={\frac {5-2x}{2x^{2}}}-{\frac {2x}{2x^{2}}}\\&=\mathbf {\frac {5-4x}{2x^{2}}} \end{aligned}}}
8. ${\displaystyle \int (2x-1)e^{-3x+1}dx}$
解答：令${\displaystyle u=2x-1}$${\displaystyle dv=e^{-3x+1}dx}$
${\displaystyle du=2dx}$${\displaystyle v=\int e^{-3x+1}dx}$
欲求${\displaystyle v}$，令${\displaystyle w=-3x+1}$${\displaystyle dw=-3dx}$${\displaystyle dx={\frac {-dw}{3}}}$，則
${\displaystyle v=\int e^{-3x+1}dx=\int e^{w}({\frac {-1}{3}})dw={\frac {-e^{w}}{3}}={\frac {-e^{-3x+1}}{3}}}$，故
{\displaystyle {\begin{aligned}\int (2x-1)e^{-3x+1}dx&=\int udv\\&=uv-\int vdu\\&=(2x-1){\frac {-e^{-3x+1}}{3}}-\int {\frac {-e^{-3x+1}}{3}}(2)dx\\&={\frac {(1-2x)e^{-3x+1}}{3}}+{\frac {2}{3}}\int e^{-3x+1}dx\\&={\frac {(1-2x)e^{-3x+1}}{3}}+{\frac {2}{3}}\int {\frac {-e^{w}}{3}}dw\\&={\frac {3(1-2x)e^{-3x+1}}{9}}-{\frac {2}{9}}e^{w}\\&={\frac {(3-6x)e^{-3x+1}}{9}}-{\frac {2}{9}}e^{-3x+1}\\&=\mathbf {\frac {(1-6x)e^{-3x+1}}{9}} \end{aligned}}}