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#<math>\int\frac{3x}{2}dx</math><br>解答：欲找到函数<math>F</math>，使<br><math>F'(x)=\frac{3x}{2}</math><br>由于<br><math>\frac{d}{dx}x^2=2x</math><br>因此须找到常数<math>a</math>，使<br><math>\frac{d}{dx}ax^2=2ax=\frac{3x}{2}</math><br>解出<math>a</math>，得<br><math>2ax=\frac{3x}{2}\implies a=\frac{3}{4}</math><br>故<br><math>\int\frac{3x}{2}=\frac{3}{4}x^2+C</math><br>求导验证：<br><math>\frac{d}{dx}\left(\frac{3}{4}x^{2}+C\right)=\frac{3}{4}(2x)=\frac{3x}{2}</math><br><br>
#求<math>f(x)=2x^4</math>的原函数<br>解答：由于<br><math>\frac{d}{dx}x^{5}=5x^{4}</math><br>须找到常数<math>a</math>，使<br><math>\frac{d}{dx}ax^{5}=5ax^{4}=2x^{4}</math><br>解出<math>a</math>，得<br><math>5ax^{4}=2x^{4}\implies a=\frac{2}{5}</math><br>故原函数为<br><math>\mathbf{\frac{2}{5}x^{5}+C}</math><br>求导验证：<br><math>\frac{d}{dx}\int 2x^4 dx=\frac{d}{dx}(\frac{2}{5}x^{5}+C)=\frac{2}{5}(5x^{4})=2x^{4}</math><br><br>
#<math>\int(7x^2+3\cos x-e^x)dx</math><br>解答：<br><math>\begin{align}\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{align}</math><br><br>
#<math>\int(\frac{2}{5x}+\sin x)dx</math><br>解答：<br><math>\begin{align}\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{align}</math><br><br>
#<math>\int x\sin 2x^2dx</math><br>解答：令<math>u=2x^{2}</math>，<math>du=4xdx</math>，<math>dx=\frac{du}{4x}</math><br><math>\begin{align}\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{align}</math>
#<math>\int-3e^{\sin x}\cos xdx</math><br>解答：令<math>u=\sin x</math>，<math>du=\cos xdx</math>，则<math>dx=\frac{du}{\cos x}</math><br><math>\begin{align}\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{align}</math>
#<math>\int \frac{2x-5}{x^3}dx</math><br>解答：令<math>u=2(x-5)</math>，<math>v=\int\frac{dx}{x^{3}}=-\frac{1}{2x^{2}}</math><br><math>\begin{align}\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{align}</math>
#<math>\int(2x-1)e^{-3x+1}dx</math><br>解答：令<math>u=2x-1</math>，<math>dv=e^{-3x+1}dx</math><br>则<math>du=2dx</math>，<math>v=\int e^{-3x+1}dx</math><br>欲求<math>v</math>，令<math>w=-3x+1</math>，<math>dw=-3dx</math>，<math>dx=\frac{-dw}{3}</math>，则<br><math>v=\int e^{-3x+1}dx=\int e^{w}(\frac{-1}{3})dw=\frac{-e^{w}}{3}=\frac{-e^{-3x+1}}{3}</math>，故<br><math>\begin{align}\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{align}</math>

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