微積分學/不定積分/練習答案

1. $\int {\frac {3x}{2}}dx$
   解答：欲找到函數$F$，使
   $F'(x)={\frac {3x}{2}}$
   由於
   ${\frac {d}{dx}}x^{2}=2x$
   因此須找到常數$a$，使
   ${\frac {d}{dx}}ax^{2}=2ax={\frac {3x}{2}}$
   解出$a$，得
   $2ax={\frac {3x}{2}}\implies a={\frac {3}{4}}$
   故
   $\int {\frac {3x}{2}}={\frac {3}{4}}x^{2}+C$
   求導驗證：
   ${\frac {d}{dx}}\left({\frac {3}{4}}x^{2}+C\right)={\frac {3}{4}}(2x)={\frac {3x}{2}}$
   
   
2. 求$f(x)=2x^{4}$的原函數
   解答：由於
   ${\frac {d}{dx}}x^{5}=5x^{4}$
   須找到常數$a$，使
   ${\frac {d}{dx}}ax^{5}=5ax^{4}=2x^{4}$
   解出$a$，得
   $5ax^{4}=2x^{4}\implies a={\frac {2}{5}}$
   故原函數為
   $\mathbf {{\frac {2}{5}}x^{5}+C}$
   求導驗證：
   ${\frac {d}{dx}}\int 2x^{4}dx={\frac {d}{dx}}({\frac {2}{5}}x^{5}+C)={\frac {2}{5}}(5x^{4})=2x^{4}$
   
   
3. $\int (7x^{2}+3\cos x-e^{x})dx$
   解答：
   ${\begin{aligned}\int (7x^{2}+3\cos x-e^{x})dx&=7\int x^{2}dx+3\int \cos xdx-\int e^{x}dx\\&=7({\frac {x^{3}}{3}})+3\sin x-e^{x}+C\\&=\mathbf {{\frac {7}{3}}x^{3}+3\sin x-e^{x}+C} \end{aligned}}$
   
   
4. $\int ({\frac {2}{5x}}+\sin x)dx$
   解答：
   ${\begin{aligned}\int ({\frac {2}{5x}}+\sin x)dx&={\frac {2}{5}}\int {\frac {dx}{x}}+\int \sin xdx\\&=\mathbf {{\frac {2}{5}}\ln |x|-\cos x+C} \end{aligned}}$
   
   
5. $\int x\sin 2x^{2}dx$
   解答：令$u=2x^{2}$，$du=4xdx$，$dx={\frac {du}{4x}}$
   ${\begin{aligned}\int x\sin 2x^{2}dx&=\int x\sin u{\frac {du}{4x}}\\&={\frac {1}{4}}\int \sin udu\\&=-{\frac {\cos u}{4}}+C\\&=-\mathbf {{\frac {\cos 2x^{2}}{4}}+C} \end{aligned}}$
6. $\int -3e^{\sin x}\cos xdx$
   解答：令$u=\sin x$，$du=\cos xdx$，則$dx={\frac {du}{\cos x}}$
   ${\begin{aligned}\int -3e^{\sin x}\cos xdx&=-3\int e^{u}\cos x{\frac {du}{\cos x}}\\&=-3\int e^{u}du\\&=-3e^{u}+C\\&=\mathbf {-3e^{\sin x}+C} \end{aligned}}$
7. $\int {\frac {2x-5}{x^{3}}}dx$
   解答：令$u=2(x-5)$，$v=\int {\frac {dx}{x^{3}}}=-{\frac {1}{2x^{2}}}$
   ${\begin{aligned}\int {\frac {2x-5}{x^{3}}}dx&=\int udv\\&=uv-\int vdu\\&=(2x-5)(-{\frac {1}{2x^{2}}})-\int (-{\frac {1}{2x^{2}}})2dx\\&={\frac {5-2x}{2x^{2}}}+\int {\frac {dx}{x^{2}}}\\&={\frac {5-2x}{2x^{2}}}-{\frac {1}{x}}\\&={\frac {5-2x}{2x^{2}}}-{\frac {2x}{2x^{2}}}\\&=\mathbf {\frac {5-4x}{2x^{2}}} \end{aligned}}$
8. $\int (2x-1)e^{-3x+1}dx$
   解答：令$u=2x-1$，$dv=e^{-3x+1}dx$
   則$du=2dx$，$v=\int e^{-3x+1}dx$
   欲求$v$，令$w=-3x+1$，$dw=-3dx$，$dx={\frac {-dw}{3}}$，則
   $v=\int e^{-3x+1}dx=\int e^{w}({\frac {-1}{3}})dw={\frac {-e^{w}}{3}}={\frac {-e^{-3x+1}}{3}}$，故
   ${\begin{aligned}\int (2x-1)e^{-3x+1}dx&=\int udv\\&=uv-\int vdu\\&=(2x-1){\frac {-e^{-3x+1}}{3}}-\int {\frac {-e^{-3x+1}}{3}}(2)dx\\&={\frac {(1-2x)e^{-3x+1}}{3}}+{\frac {2}{3}}\int e^{-3x+1}dx\\&={\frac {(1-2x)e^{-3x+1}}{3}}+{\frac {2}{3}}\int {\frac {-e^{w}}{3}}dw\\&={\frac {3(1-2x)e^{-3x+1}}{9}}-{\frac {2}{9}}e^{w}\\&={\frac {(3-6x)e^{-3x+1}}{9}}-{\frac {2}{9}}e^{-3x+1}\\&=\mathbf {\frac {(1-6x)e^{-3x+1}}{9}} \end{aligned}}$