📝 题目
3. 1.13 求下列不定积分:
(1) $\displaystyle \int \frac{\mathrm{d}x}{{\left( 1 + \cos x\right) }^{2}}$ ; (2) $\displaystyle \int \frac{\mathrm{d}\theta }{1 + {r}^{2} - {2r}\cos \theta }\left( {0 < r < 1}\right)$ ；
(3) $\displaystyle{\int \frac{\sqrt{x}}{1 + \sqrt[4]{{x}^{3}}}\mathrm{\;d}x}$ ； (4) $\displaystyle{\int \frac{\mathrm{d}x}{1 + \sqrt{x} + \sqrt{x + 1}}}$ ;
(5) $\displaystyle{\int \frac{x}{x + \sqrt{{x}^{2} - 1}}\mathrm{\;d}x}$ ; (6) $\displaystyle{\int \frac{\mathrm{d}x}{1 + \sqrt{1 - {2x} - {x}^{2}}}}$ .
💡 答案与解析
### 3.1.13 求下列不定积分
#### (1) $\displaystyle \int \frac{\mathrm{d}x}{{\left( 1 + \cos x\right) }^{2}}$
**解答步骤：**
利用半角公式 $1 + \cos x = 2\cos^2\frac{x}{2}$，则 $$ \frac{1}{(1+\cos x)^2} = \frac{1}{4\cos^4\frac{x}{2}} $$ 于是 $$ \int \frac{dx}{(1+\cos x)^2} = \frac14 \int \sec^4\frac{x}{2} \, dx $$ 令 $t = \frac{x}{2}$，则 $dx = 2\,dt$，原积分化为 $$ \frac14 \int \sec^4 t \cdot 2\,dt = \frac12 \int \sec^4 t \, dt $$ 利用 $\sec^4 t = (1+\tan^2 t)\sec^2 t$，令 $u = \tan t$，则 $du = \sec^2 t\, dt$， $$ \frac12 \int (1+u^2)\, du = \frac12\left(u + \frac{u^3}{3}\right) + C $$ 代回 $u = \tan t = \tan\frac{x}{2}$，得 $$ \boxed{\frac12 \tan\frac{x}{2} + \frac16 \tan^3\frac{x}{2} + C} $$
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#### (2) $\displaystyle \int \frac{\mathrm{d}\theta }{1 + {r}^{2} - {2r}\cos \theta }\left( {0 < r < 1}\right)$
**解答步骤：**
这是典型的用万能代换 $t = \tan\frac{\theta}{2}$ 的积分。 由 $\cos\theta = \frac{1-t^2}{1+t^2}$，$d\theta = \frac{2}{1+t^2} dt$，分母变为 $$ 1+r^2 - 2r\frac{1-t^2}{1+t^2} = \frac{(1+r^2)(1+t^2) - 2r(1-t^2)}{1+t^2} $$ 分子化简： $$ (1+r^2)(1+t^2) - 2r + 2r t^2 = (1+r^2 - 2r) + (1+r^2+2r)t^2 = (1-r)^2 + (1+r)^2 t^2 $$ 因此积分化为 $$ \int \frac{\frac{2}{1+t^2}\, dt}{\frac{(1-r)^2 + (1+r)^2 t^2}{1+t^2}} = 2\int \frac{dt}{(1-r)^2 + (1+r)^2 t^2} $$ 提取 $(1-r)^2$： $$ = \frac{2}{(1-r)^2} \int \frac{dt}{1 + \left(\frac{1+r}{1-r}t\right)^2} $$ 令 $u = \frac{1+r}{1-r}t$，则 $dt = \frac{1-r}{1+r} du$，得 $$ \frac{2}{(1-r)^2} \cdot \frac{1-r}{1+r} \int \frac{du}{1+u^2} = \frac{2}{(1-r)(1+r)} \arctan u + C $$ 代回 $u = \frac{1+r}{1-r}\tan\frac{\theta}{2}$，得 $$ \boxed{\frac{2}{1-r^2} \arctan\left( \frac{1+r}{1-r} \tan\frac{\theta}{2} \right) + C} $$
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#### (3) $\displaystyle{\int \frac{\sqrt{x}}{1 + \sqrt[4]{{x}^{3}}}\mathrm{\;d}x}$
**解答步骤：**
令 $x = t^4$，则 $dx = 4t^3 dt$，$\sqrt{x} = t^2$，$\sqrt[4]{x^3} = t^3$，积分变为 $$ \int \frac{t^2}{1+t^3} \cdot 4t^3 dt = 4\int \frac{t^5}{1+t^3} dt $$ 做多项式除法：$t^5 = t^2(t^3+1) - t^2$，所以 $$ \frac{t^5}{1+t^3} = t^2 - \frac{t^2}{1+t^3} $$ 于是 $$ 4\int \left(t^2 - \frac{t^2}{1+t^3}\right) dt = 4\left( \frac{t^3}{3} - \frac13 \ln|1+t^3| \right) + C $$ 代回 $t = x^{1/4}$，得 $$ \boxed{\frac{4}{3} x^{3/4} - \frac{4}{3} \ln\left(1 + x^{3/4}\right) + C} $$
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#### (4) $\displaystyle{\int \frac{\mathrm{d}x}{1 + \sqrt{x} + \sqrt{x + 1}}}$
**解答步骤：**
有理化分母： $$ \frac{1}{1+\sqrt{x}+\sqrt{x+1}} = \frac{1+\sqrt{x} - \sqrt{x+1}}{(1+\sqrt{x}+\sqrt{x+1})(1+\sqrt{x}-\sqrt{x+1})} $$ 分母为 $(1+\sqrt{x})^2 - (x+1) = 1 + 2\sqrt{x} + x - x - 1 = 2\sqrt{x}$ 所以原积分等于 $$ \int \frac{1+\sqrt{x} - \sqrt{x+1}}{2\sqrt{x}} dx = \frac12 \int \left( \frac{1}{\sqrt{x}} + 1 - \frac{\sqrt{x+1}}{\sqrt{x}} \right) dx $$ 前两项积分：$\frac12(2\sqrt{x} + x)$。第三项：令 $t = \sqrt{x}$，则 $x = t^2$，$dx = 2t dt$， $$ \int \frac{\sqrt{x+1}}{\sqrt{x}} dx = \int \frac{\sqrt{t^2+1}}{t} \cdot 2t dt = 2\int \sqrt{t^2+1}\, dt $$ 而 $\displaystyle{\int \sqrt{t^2+1}\, dt = \frac{t}{2}\sqrt{t^2+1} + \frac12 \ln|t+\sqrt{t^2+1}| + C}$，代回 $t=\sqrt{x}$，得 $$ \sqrt{x}\sqrt{x+1} + \ln(\sqrt{x}+\sqrt{x+1}) $$ 因此原积分为 $$ \frac12\left(2\sqrt{x} + x - \sqrt{x}\sqrt{x+1} - \ln(\sqrt{x}+\sqrt{x+1})\right) + C $$ 即 $$ \boxed{\sqrt{x} + \frac{x}{2} - \frac12\sqrt{x(x+1)} - \frac12\ln(\sqrt{x}+\sqrt{x+1}) + C} $$
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#### (5) $\displaystyle{\int \frac{x}{x + \sqrt{{x}^{2} - 1}}\mathrm{\;d}x}$
**解答步骤：**
有理化分母： $$ \frac{x}{x+\sqrt{x^2-1}} = \frac{x(x-\sqrt{x^2-1})}{x^2 - (x^2-1)} = x(x - \sqrt{x^2-1}) $$ 所以积分化为 $$ \int (x^2 - x\sqrt{x^2-1})\, dx = \frac{x^3}{3} - \int x\sqrt{x^2-1}\, dx $$ 对于第二项，令 $u = x^2-1$，则 $du = 2x dx$， $$ \int x\sqrt{x^2-1}\, dx = \frac12 \int \sqrt{u}\, du = \frac12 \cdot \frac{2}{3} u^{3/2} = \frac13 (x^2-1)^{3/2} $$ 因此结果为 $$ \boxed{\frac{x^3}{3} - \frac13 (x^2-1)^{3/2} + C} $$
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#### (6) $\displaystyle{\int \frac{\mathrm{d}x}{1 + \sqrt{1 - {2x} - {x}^{2}}}}$
**解答步骤：**
先配方：$1 - 2x - x^2 = 2 - (x+1)^2$，令 $x+1 = \sqrt{2}\sin t$，则 $dx = \sqrt{2}\cos t\, dt$，且 $$ \sqrt{1-2x-x^2} = \sqrt{2}\cos t $$ 分母变为 $1 + \sqrt{