《高等数学》习题2.5选做

求下列不定积分： 1. $\displaystyle \int \left(\frac{a}{\sqrt{x}} -\frac{b}{x^{2}} +3c\sqrt[3]{x^{2}}\right)\mathrm{d} x$ 解：$\displaystyle =2a\sqrt{x} +\frac{b}{x} +\frac{9}{5} cx^{5/3} +C$ 2. $\displaystyle \int \left( 1+\sqrt{x}\right)^{2}\mathrm{d} x$ 解：$\displaystyle =\int \left( 1+2\sqrt{x} +x\right)\mathrm{d} x=x+\frac{4}{3} x^{3/2} +\frac{1}{2} x^{2} +C$ 3. $\displaystyle \int a\sec^{2} t\ \mathrm{d} t$ 解：$\displaystyle =a\int \sec^{2} t\ \mathrm{d} t=a\tan t+C$ 4. $\displaystyle \int \tan^{2} t\ \mathrm{d} t$ 解：$\displaystyle =\int \left(\sec^{2} t-1\right)\mathrm{d} t=\tan t-t+C$ 5. $\displaystyle \int \cot^{2} \varphi \ \mathrm{d} \varphi $ 解：$\displaystyle =\int \left(\csc^{2} \varphi -1\right)\mathrm{d} \varphi =-\cot \varphi -\varphi +C$ 6. $\displaystyle \int \frac{x^{2} +3}{1+x^{2}}\mathrm{d} x$ 解：$\displaystyle =\int \left( 1+\frac{2}{1+x^{2}}\right)\mathrm{d} x=x+2\arctan x+C$ 7. $\displaystyle \int \left(\frac{3}{\sqrt{x}} +\frac{4}{\sqrt{1-x^{2}}}\right)\mathrm{d} x$ 解：$\displaystyle =6\sqrt{x} +4\arcsin x+C$ 12. $\displaystyle \int ( 2\cosh x-\sinh x)\mathrm{d} x$ 解：$\displaystyle =\int \left(\mathrm{e}^{x} +\mathrm{e}^{-x} -\frac{\mathrm{e}^{x} -\mathrm{e}^{-x}}{2}\right)\mathrm{d} x=\int \left(\frac{1}{2}\mathrm{e}^{x} +\frac{3}{2}\mathrm{e}^{-x}\right)\mathrm{d} x=\frac{1}{2}\mathrm{e}^{x} -\frac{3}{2}\mathrm{e}^{-x} +C$ 14. $\displaystyle \int \frac{1}{\sin^{2} x\cos^{2} x}\mathrm{d} x$ 解：$\displaystyle =\int \frac{\sin^{2} x+\cos^{2} x}{\sin^{2} x\cos^{2} x}\mathrm{d} x=\int (\sec x+\csc x)\mathrm{d} x=\tan x-\cot x+C$ 15. $\displaystyle \int \frac{2^{x+1} +3^{x-2}}{6^{x}}\mathrm{d} x$ 解：$\displaystyle =\int \left[ 2\cdot \left(\frac{1}{3}\right)^{x} +\frac{1}{9} \cdot \left(\frac{1}{2}\right)^{x}\right]\mathrm{d} x=-\frac{2}{\ln 3} \cdot \left(\frac{1}{3}\right)^{x} -\frac{1}{9\ln 2} \cdot \left(\frac{1}{2}\right)^{x} +C$ 16. $\displaystyle \int \frac{1}{x^{2}\left( 1+x^{2}\right)}\mathrm{d} x$ 解：$\displaystyle =\int \left[\frac{1}{x^{2}} -\frac{1}{1+x^{2}}\right]\mathrm{d} x=-\frac{1}{x} -\arctan x+C$ 17. 求解微分方程 $\displaystyle y''( x) =a+b\mathrm{e}^{-x}$ 解：$\displaystyle y'( x) =\int \left( a+b\mathrm{e}^{-x}\right)\mathrm{d} x=ax+C_{1} -b\mathrm{e}^{-x}$，也就有 $\displaystyle y( x) =\int y'( x) \ \mathrm{d} x=\frac{a}{2} x^{2} +C_{1} x+C_{2} +b\mathrm{e}^{-x}$。 18. 设 $\displaystyle f( x)$ 满足下列方程 $\displaystyle xf'( x) +f( x) =x^{3} +1$，求 $\displaystyle f( x)$ 解：注意到 $\displaystyle ( xf) '=f+xf'$，因此 $\displaystyle xf( x) =\int \left( x^{3} +1\right)\mathrm{d} x=\frac{1}{4} x^{4} +x+C$，也即 $\displaystyle f( x) =\frac{1}{4} x^{3} +1+\frac{C}{x}$。