微积分学/不定积分/练习答案

1. $\int \frac{3x}{2}dx$
   解答：欲找到函数$F$，使
   $F^{\prime }(x)=\frac{3x}{2}$
   由于
   $\frac{d}{dx}{x}^2=2x$
   因此须找到常数$a$，使
   $\frac{d}{dx}a{x}^2=2ax=\frac{3x}{2}$
   解出$a$，得
   $2ax=\frac{3x}{2}⟹a=\frac{3}{4}$
   故
   $\int \frac{3x}{2}=\frac{3}{4}{x}^2+C$
   求导验证：
   $\frac{d}{dx}(\frac{3}{4}{x}^2+C)=\frac{3}{4}(2x)=\frac{3x}{2}$
   
   
2. 求$f(x)=2x^4$的原函数
   解答：由于
   $\frac{d}{dx}{x}^5=5x^4$
   须找到常数$a$，使
   $\frac{d}{dx}a{x}^5=5a{x}^4=2x^4$
   解出$a$，得
   $5a{x}^4=2x^4⟹a=\frac{2}{5}$
   故原函数为
   $\frac{2}{5}{𝐱}^{\mathrm{𝟓}}\mathbf{+}𝐂$
   求导验证：
   $\frac{d}{dx}\int 2x^{4}dx=\frac{d}{dx}(\frac{2}{5}{x}^5+C)=\frac{2}{5}(5x^4)=2x^4$
   
   
3. $\int (7x^2+3\cos x-e^x)dx$
   解答：
   $\begin{array}{cc}\int (7x^2+3\cos x-e^x)dx& =7\int x^{2}dx+3\int \cos xdx-\int e^{x}dx\\ & =7(\frac{x^3}{3})+3\sin x-e^x+C\\ & =\frac{7}{3}{𝐱}^{\mathrm{𝟑}}\mathbf{+}\mathrm{𝟑}\sin 𝐱\mathbf{-}{𝐞}^{𝐱}\mathbf{+}𝐂\end{array}$
   
   
4. $\int (\frac{2}{5x}+\sin x)dx$
   解答：
   $\begin{array}{cc}\int (\frac{2}{5x}+\sin x)dx& =\frac{2}{5}\int \frac{dx}{x}+\int \sin xdx\\ & =\frac{2}{5}\ln \mathbf{|}𝐱\mathbf{|}\mathbf{-}\cos 𝐱\mathbf{+}𝐂\end{array}$
   
   
5. $\int x\sin 2x^{2}dx$
   解答：令$u=2x^2$，$du=4xdx$，$dx=\frac{du}{4x}$
   $\begin{array}{cc}\int x\sin 2x^{2}dx& =\int x\sin u\frac{du}{4x}\\ & =\frac{1}{4}\int \sin udu\\ & =-\frac{\cos u}{4}+C\\ & =-\frac{\cos 2x^2}{4}\mathbf{+}𝐂\end{array}$
6. $\int -3e^{\sin x}\cos xdx$
   解答：令$u=\sin x$，$du=\cos xdx$，则$dx=\frac{du}{\cos x}$
   $\begin{array}{cc}\int -3e^{\sin x}\cos xdx& =-3\int e^u\cos x\frac{du}{\cos x}\\ & =-3\int e^{u}du\\ & =-3e^u+C\\ & =\mathbf{-}\mathrm{𝟑}{𝐞}^{\sin 𝐱}\mathbf{+}𝐂\end{array}$
7. $\int \frac{2x-5}{x^3}dx$
   解答：令$u=2(x-5)$，$v=\int \frac{dx}{x^3}=-\frac{1}{2x^2}$
   $\begin{array}{cc}\int \frac{2x-5}{x^3}dx& =\int udv\\ & =uv-\int vdu\\ & =(2x-5)(-\frac{1}{2x^2})-\int (-\frac{1}{2x^2})2dx\\ & =\frac{5-2x}{2x^2}+\int \frac{dx}{x^2}\\ & =\frac{5-2x}{2x^2}-\frac{1}{x}\\ & =\frac{5-2x}{2x^2}-\frac{2x}{2x^2}\\ & =\frac{5-4x}{2x^2}\end{array}$
8. $\int (2x-1)e^{-3x+1}dx$
   解答：令$u=2x-1$，$dv=e^{-3x+1}dx$
   则$du=2dx$，$v=\int e^{-3x+1}dx$
   欲求$v$，令$w=-3x+1$，$dw=-3dx$，$dx=\frac{-dw}{3}$，则
   $v=\int e^{-3x+1}dx=\int e^w(\frac{-1}{3})dw=\frac{-e^w}{3}=\frac{-e^{-3x+1}}{3}$，故
   $\begin{array}{cc}\int (2x-1)e^{-3x+1}dx& =\int udv\\ & =uv-\int vdu\\ & =(2x-1)\frac{-e^{-3x+1}}{3}-\int \frac{-e^{-3x+1}}{3}(2)dx\\ & =\frac{(1-2x)e^{-3x+1}}{3}+\frac{2}{3}\int e^{-3x+1}dx\\ & =\frac{(1-2x)e^{-3x+1}}{3}+\frac{2}{3}\int \frac{-e^w}{3}dw\\ & =\frac{3(1-2x)e^{-3x+1}}{9}-\frac{2}{9}{e}^w\\ & =\frac{(3-6x)e^{-3x+1}}{9}-\frac{2}{9}{e}^{-3x+1}\\ & =\frac{(1-6x)e^{-3x+1}}{9}\end{array}$