中册 4.1 不定积分计算 第23题

解题过程： （1）令 $t=\sqrt[4]{x}$ ，则 $$ \begin{aligned} \int \frac{1}{\sqrt{x}(1+\sqrt[4]{x})^{3}} \mathrm{~d} x & =\int \frac{4 t^{3}}{t^{2}(1+t)^{3}} \mathrm{~d} t=4 \int\left[\frac{1}{(1+t)^{2}}-\frac{1}{(1+t)^{3}}\right] \mathrm{d} t \\ & =\frac{-4}{1+t}+\frac{2}{(1+t)^{2}}+C=-\frac{4}{1+\sqrt[4]{x}}+\frac{2}{(1+\sqrt[4]{x})^{2}}+C \end{aligned} $$ （2）令 $\displaystyle t=\sqrt{\frac{1-x}{1+x}}$ ，则 $\displaystyle x=\frac{1-t^{2}}{1+t^{2}}, \mathrm{~d} x=\frac{-4 t \mathrm{~d} t}{\left(1+t^{2}\right)^{2}}$ 。于是 $$ \begin{aligned} \int \frac{1}{x^{2}} \sqrt{\frac{1-x}{1+x}} \mathrm{~d} x & =\int\left(\frac{1+t^{2}}{1-t^{2}}\right)^{2} \cdot \frac{1}{t} \cdot \frac{-4 t}{\left(1+t^{2}\right)^{2}} \mathrm{~d} t=\int \frac{-4}{\left(1-t^{2}\right)^{2}} \mathrm{~d} t=-\int\left[\frac{1}{(1+t)^{2}}+\frac{1}{(1-t)^{2}}+\frac{2}{1-t^{2}}\right] \mathrm{d} t \\ & =-2 \int \frac{1}{1-t^{2}} \mathrm{~d} t-\int\left[\frac{1}{(1+t)^{2}}+\frac{1}{(1-t)^{2}}\right] \mathrm{d} t=-\ln \left|\frac{1+t}{1-t}\right|-\frac{1}{1-t}+\frac{1}{1+t}+C \\ & =-\ln \left|\frac{1+\sqrt{1-x^{2}}}{x}\right|-\frac{\sqrt{1-x^{2}}}{x}+C \end{aligned} $$ （3）令 $\displaystyle t=\sqrt{\frac{1+x}{2-x}}$ ，则 $\displaystyle x=\frac{2 t^{2}-1}{1+t^{2}}, \mathrm{~d} x=\frac{6 t \mathrm{~d} t}{\left(1+t^{2}\right)^{2}}$ ．于是 $$ \begin{aligned} \int \frac{\sqrt{2+x-x^{2}}}{x} \mathrm{~d} x & =\int \frac{2-x}{x} \sqrt{\frac{1+x}{2-x}} \mathrm{~d} x=\int \frac{18 t^{2}}{\left(2 t^{2}-1\right)\left(1+t^{2}\right)^{2}} \mathrm{~d} t \\ & =6 \int \frac{1}{\left(1+t^{2}\right)^{2}} \mathrm{~d} t-2 \int \frac{1}{1+t^{2}} \mathrm{~d} t+4 \int \frac{1}{2 t^{2}-1} \mathrm{~d} t \\ & =3\left(\arctan t+\frac{t}{1+t^{2}}\right)-2 \arctan t+\sqrt{2} \ln \left|\frac{\sqrt{2} t-1}{\sqrt{2} t+1}\right|+C \\ & =\arctan t+\frac{3 t}{1+t^{2}}+\sqrt{2} \ln \left|\frac{\sqrt{2} t-1}{\sqrt{2} t+1}\right|+C \\ & =\arctan \sqrt{\frac{1+x}{2-x}}+\sqrt{2+x-x^{2}}+\ln \left|\frac{\sqrt{2+2 x}-\sqrt{2-x}}{\sqrt{2+2 x}+\sqrt{2-x}}\right|+C \end{aligned} $$ （4）令 $\displaystyle \sqrt[3]{\frac{x-1}{x+1}}=t$ ，则 $\displaystyle x=\frac{1+t^{3}}{1-t^{3}}, \mathrm{~d} x=\frac{6 t^{2}}{\left(1-t^{3}\right)^{2}} \mathrm{~d} t$ 。于是 $$ \int \frac{\mathrm{d} x}{\sqrt[3]{(x+1)^{2}(x-1)^{4}}}=\frac{3}{2} \int \frac{\mathrm{~d} t}{t^{2}}=-\frac{3}{2 t}+C=-\frac{3}{2} \sqrt[3]{\frac{x+1}{x-1}}+C $$ （5）令 $t=\sqrt{1+2 x}$ ，则 $\displaystyle \int \frac{x+2}{\sqrt{1+2 x}} \mathrm{~d} x=\frac{1}{2} \int\left(t^{2}+3\right) \mathrm{d} t=\frac{1}{2}\left(\frac{1}{3} t^{3}+3 t\right)+C=\frac{1}{6} \sqrt{(1+2 x)^{3}}+\frac{3}{2} \sqrt{1+2 x}+C$ ．