高数 06.03 积分习题课01不定积分

$\color{blue}{一元函数积分学习题课}$

$\color{blue}{不定积分}$

一、考试内容
原函数和不定积分的概念，不定积分的基本性质，不定积分的基本公式，不定积分的换元积分法和分部积分法。
二、考试要求
1.理解原函数与不定积分的概念，掌握不定积分的基本性质
2.熟练掌握不定积分的基本公式
3.熟练掌握不定积分的第一类换元法，掌握不定积分的第二类换元法(三角代换和简单的根式代换)
4.熟练掌握不定积分的分部积分法
三、基本知识
(一)原函数与不定积分的概念
$1.定义 若在区间I上定义的两个函数F(x)及f(x) \\ 满足F^{\prime}(x) = f(x)或dF(x) = f(x) dx,则称F(x)\\ 为f(x)在区间I上的一个原函数.$
$2.若F(x)是f(x)的一个原函数,则f(x)的所有\\ 原函数为F(x) + C(C为任意常数).$
$3.f(x)在区间I上的原函数的全体称为f(x)在I上\\ 的不定积分,记作\int{f(x)dx},其中$
$\int –积分号; \quad f(x) –被积函数; \\ x–积分变量; \quad f(x)dx –被积表达式.$
$若F^{\prime}(x) = f(x),则\\ \int{f(x)dx} = F(x) + C (C为任意常数)$
(二)不定积分的性质
$(1) \dfrac{d}{dx}[\int{f(x)dx}] = f(x)或d[\int{f(x)dx}] = f(x)dx$
$(2)\int{F^{\prime}(x) dx} = F(x) + C 或 \int{dF(x)} = F(x) + C$
$(3) \int{kf(x)dx} = k\int{f(x)dx}$
$(4)\int{[f(x) \pm g(x)] dx} = \int{f(x)dx} \pm \int{g(x)dx}$

(三)不定积分的基本公式
$(1) \int{kdx} = kx + C$
$(2) \int{x^{\mu} dx} = \dfrac{1}{\mu + 1} x^{\mu + 1} + C (\mu \neq -1)$
$(3) \int{\dfrac{dx}{x}} = \ln|x| + C$
$(4) \int{\dfrac{dx}{1 + x^2}} = \arctan{x} + C$
$(5) \int{\dfrac{dx}{\sqrt{1 - x^2}}} = \arcsin{x} + C$
$(6) \int{\cos{x}dx} = \sin{x} + C$
$(7) \int{\sin{x}dx} = -\cos{x} + C$
$(8) \int{\dfrac{dx}{\cos^2{x}}} = \int{sec^2{x} dx} = \tan{x} + C$
$(9) \int{\dfrac{dx}{\sin^2{x}}} = \int{csc^2{x} dx} = -\cot{x} + C$
$(10) \int{\sec{x}\tan{x} dx} = \sec{x} + C$
$(11) \int{\csc{x} \cot{x} dx} = -\csc{x} + C$
$(12) \int{e^x dx} = e^x + C$
$(13) \int{a^x dx} = \dfrac{a^x}{\ln{a}} + C$

(四)基本积分方法
1.直接积分法
2.第一类换元积分公式
$\int{f(u(x))u^{\prime}(x) dx} = \int{f(u)du} = F(u) + C$
3.第二类换元积分公式(变量代换)
4.分部积分法
$\int{u(x) dv(x)} = u(x)v(x) - \int{v(x) du(x)}$

例题分析
(1)单项选择题
$例1 \sqrt{x}是({\color{blue}{B} })的一个原函数.$
$A.\dfrac{1}{2x} \quad B.\dfrac{1}{2\sqrt{x}} \quad C.\ln{x} \quad D.\sqrt{x^3}$
$解:(\sqrt{x})^{\prime} = \dfrac{1}{2} x^{-\frac{1}{2}} = \dfrac{1}{2\sqrt{x}} \ 选B$

$例2.({\color{blue}{D} })是函数f(x) = \dfrac{1}{2x}的原函数.$
$A. F(x) = \ln{2x} \quad B.F(x) = -\dfrac{1}{2x^2} \\ C.F(x) = \ln{2 + x} \quad D.F(x) = \dfrac{1}{2}\ln{3x}$
$解:F(x) = \int{\dfrac{1}{2x} dx} \\ = \dfrac{1}{2} \int{\dfrac{1}{x} dx} \\ = \dfrac{1}{2} \int{ d \ln{x}} \\ = \dfrac{1}{2} \ln{x} + C = \dfrac{1}{2} \ln{3x} - \dfrac{1}{2} \ln{3} + C$

$例3.下列等式中,({\color{blue}{D}})是正确的.$
$A. \int{f^{\prime}(x)dx} = f(x) \\ B.\int{f^{\prime}(e^x)dx} = f(e^x) + C \\ C.\int{f^{\prime}(\sqrt{x}) dx} = 2\sqrt{x}f(\sqrt{x}) + C \\ D.\int{xf^{\prime}(1 - x^2) dx} = -\dfrac{1}{2}f(1 - x^2) + C$
$解:\\ A. \int{f^{\prime}(x)dx} = f(x) + C \\ B. [f(e^x) + C]^{\prime} = e^xf^{\prime}(e^x) \\ C.[2\sqrt{x}f(\sqrt{x}) + C]^{\prime} =\\ 2 \cdot {\dfrac{1}{2\sqrt{x}}}f(\sqrt{x}) + 2\sqrt{x}f^{\prime}(\sqrt{x})\dfrac{1}{2\sqrt{x}} \\ = f^{\prime}(\sqrt{x}) + \dfrac{f(x)}{\sqrt{x}} \\ D.[-\dfrac{1}{2}f(1 - x^2) + C]^{\prime} \\ = -\dfrac{1}{2}f^{\prime}(1 - x^2) \cdot (-2x) \\ = xf^{\prime}(1 - x^2)$

$例4.若\int{f(x)dx} = F(x) + C,则\int{e^{-x}f(e^{-x})dx} = ({\color{blue}{A} })$
$A. -F(e^{-x}) + C \quad B.F(e^{-x}) + C \\ C.\dfrac{F(e^{-x})}{x} + C \quad D.F(e^x) + C$
$解:\\ \int{e^{-x}f(e^{-x})dx} \\ = -\int{f(e^{-x})d(e^{-x})} = -F(e^{-x}) + C$

$例5.下列分部积分中, u,dv选择正确的是({\color{blue}{A} })$
$A.\int{x\sin{2x}dx},令u=x, dv=sin2x dx\\ B.\int{\ln{x} dx},令u = 1, dv = \ln{x} dx \\ C.\int{x^2e^{-x} dx},令u=e^{-x}, dv = x^2dx \\ D.\int{xe^x dx},令u = e^x,dv = xdx$
$解:\\ [反对幂指三]\\ A.令u=x, dv=sin2x dx\\ \int{x\sin{2x}dx} \\ = -\dfrac{1}{2}\cos{2x} \cdot x - \int{(-\dfrac{1}{2}\cos{2x}) dx} \\ = -\dfrac{1}{2}\cos{2x} \cdot x + \dfrac{1}{4} \int{(\cos{2x}) d(2x)} \\ = -\dfrac{1}{2}\cos{2x} \cdot x + \dfrac{\sin{2x}}{4} + C\\ B.令u = \ln{x} , dv = dx\\ \int{\ln{x} dx} \\ = x \ln{x} - \int{x d\ln{x}} \\ = x \ln{x} - \int{dx} \\ = x \ln{x} - x + C \\ C.另u=x^2, dv = d(-e^{-x}) \int{x^2e^{-x} dx} \\ = - x^2 e^{-x} - \int{(-e^{-x})d(x^2)} \\ = - x^2 e^{-x} + 2\int{xe^{-x}dx} \\ = - x^2 e^{-x} + 2\int{(-x)d(e^{-x})} \\ = - x^2 e^{-x} + 2(-x)e^{-x} - \int{e^{-x} d(-x)} \\ = - x^2 e^{-x} + 2(-x)e^{-x} - e^{-x} + C \\ D.令u = x, dv = de^x \\ \int{xe^x dx} \\ = xe^x - \int{e^x dx} \\ = xe^x - e^x + C$

(二)填空题
$例6.函数f(x) = 3^x的一个原函数是\underline{\ \ \color{blue}{\dfrac{3^x}{\ln{3}}} \ \ }$
$解:\\ F(x) = \int{3^xdx} = \dfrac{3^x}{\ln{3}} + C$

$例7.函数f^{\prime}(x) 的不定积分是\underline{\ \ \color{blue}{f(x) + C } \ \ }$
$解:\int{f^{\prime}(x) dx} = f(x) + C$

$例8.设f(x) = \int{\dfrac{1}{\sqrt{1 - x}} dx}, 则f^{\prime}(0) = \underline{\ \ \color{blue}{ 1 } \ \ }$
$解:\\\ f(x) = \int{f^{\prime}(x) dx} \\ f^{\prime}(x) = \dfrac{1}{\sqrt{1 - x}} \\ f^{\prime}(x) = \dfrac{1}{\sqrt{1 - 0}} = 1$

$例9.\int{(\sin{x})^{\prime} dx} = \underline{\ \ \color{blue}{ \sin{x} + C } \ \ }$

$例10.(\int{\arcsin{x} dx})^{\prime} = \underline{\ \ \color{blue}{ \arcsin{x} } \ \ }$

$例11.d\int{e^{-x^2} dx} = \underline{\ \ \color{blue}{ e^{-x^2} dx } \ \ }$

$例12.\int{f(x) dx} =x^2e^{2x} + C, 则f(x) = \underline{\ \ \color{blue}{ 2x(x + 1)e^{2x} } \ \ }$
$f(x) = [x^2e^{2x} + C]^{\prime} \\ = 2xe^{2x} + 2x^2e^{2x} \\ =2x(x + 1)e^{2x}$

$例13.若f^{\prime}(x) = \dfrac{1}{\sqrt{1 - x^2}}, 且f(1) = \dfrac{3}{2}\pi, \\ 则f(x) = \underline{\ \ \color{blue}{ \arcsin{x} + \pi } \ \ }$
$解:\\ f(x) = \int{\dfrac{1}{\sqrt{1 - x^2}} dx} \\ = \arcsin{x} + C \\ f(1) = \arcsin{1} + C = \dfrac{3}{2}\pi \\ C = \pi \\ f(x) = \arcsin{x} + \pi$

$例14.已知f(x) = \int{(1 - 2x)^{100} dx},则f(x) = \underline{\ \ {\color{blue}{ -\dfrac{(1-2x)^{101}}{202} + C } }\ \ }$
$解:\\ f(x) = \int{(1 - 2x)^{100} dx} \\ = -\dfrac{1}{2}\int{(1 - 2x)^{100} d(1 - 2x)} \\ = -\dfrac{1}{2}\int{ d[\dfrac{1}{101}(1 - 2x)^{101}]} \\ = \dfrac{1}{202}(1-2x)^{101} + C$

$例15.\int{\dfrac{f^{\prime}(\ln{x})}{x} dx} = \underline{\ \ {\color{blue}{ f(\ln{x}) + C } }\ \ }$
$解:\\ \int{\dfrac{f^{\prime}(\ln{x})}{x} dx} \\ = \int{f^{\prime}(\ln{x}) d \ln{x}} \\ = \int{d[f(\ln{x})]} \\ = f(\ln{x}) + C$

(三)解答题
$例16.求\dfrac{1}{x}的全体原函数.$
$解:\\ \int{\dfrac{1}{x} dx} = \ln|x| + C$

$例17. 设\dfrac{1}{x}为f(x)的原函数,求f(x)$
$解:\\ f(x) = (\dfrac{1}{x})^{\prime} \\ = -x^{-2} \\ = -\dfrac{1}{x^2}$

$例18. 若f(x) = x + \sqrt{x} (x > 0),求\int{f^{\prime}(x^2) dx}$
$解:\\ f^{\prime}(x) = 1 + \dfrac{1}{2\sqrt{x}} (x > 0) \\ f^{\prime}(x^2) = 1 + \dfrac{1}{2x} \\ \int{f^{\prime}(x^2) dx} \\ = \int{(1 + \dfrac{1}{2x}) dx} \\ = x + \dfrac{1}{2} \ln{x} + C$

$例19. 已知曲线y = f(x)在点X切线的斜率为2x,且曲线过(1, 0)点,求该点曲线方程.$
$解:\\ f^{\prime}(x) = 2x \\ f(x) = \int{f^{\prime}(x) dx} \\ = \int{2x dx} \\ = x^2 + C \\ f(1) = 1 + C = 0 \\ C = -1 \\ f(x) = x^2 - 1$

$例20. 求\int{\sqrt{x}( x^2 + x^{-\frac{3}{2}}) dx}$
$解:\\ \int{\sqrt{x}( x^2 + x^{-\frac{3}{2}}) dx} \\ = \int{( x^{\frac{5}{2}} + x^{-1}) dx} \\ = \int{( x^{\frac{5}{2}} + x^{-1}) dx} \\ = \dfrac{2}{7}x^{\frac{7}{2}} + \ln{x} + C \\ = \dfrac{2}{7}x^3 \sqrt{x} + \ln{x} + C$

$例21. 求\int{e^x(3 + 2^x) dx}$
$解:\\ \int{e^x(3 + 2^x) dx} \\ = 3\int{e^x dx} + \int{e^x2^x dx} \\ = 3e^x + \int{(2e)^x dx} \\ = 3e^x + \dfrac{(2e)^x}{\ln{2e}} + C$

$例22. 求不定积分\int{x^2(x^3 - 2)dx}$
$解:\\ \int{x^2(x^3 - 2)dx} \\ = \int{(x^5 - 2x^2)dx} \\ = \dfrac{1}{6}x^6 - \dfrac{2}{3}x^3 + C \\ 解法2：\\\ \int{x^2(x^3 - 2)dx} \\ = \int{\dfrac{1}{3}(x^3 - 2)d(x^3 -2)} \\ = \dfrac{1}{3} \cdot \dfrac{1}{2}(x^2 - 2)^2 + C_1 \\ = \dfrac{1}{6}x^6 -\dfrac{4}{6}x^3 + \dfrac{4}{6} + C_1 \\ = \dfrac{1}{6}x^6 - \dfrac{2}{3}x^3 + C$

$例23. 求不定积分\int{x\sqrt{2 - 3x^2} dx}$
$解:\\ \int{x\sqrt{2 - 3x^2} dx} \\ = \int{-\dfrac{1}{6}\sqrt{2 - 3x^2} d(2 - 3x^2)} \\ = -\dfrac{1}{6} \cdot \dfrac{2}{3}\int{ d(2 - 3x^2)^{\frac{3}{2}}} \\ = -\dfrac{1}{9}(2 - 3x^2)\sqrt{2 - 3x^2} + C$

$例24. 求不定积分\int{\dfrac{1}{x} \sin{(\ln{x})} dx}$
$解:\\ \int{\dfrac{1}{x} \sin{(\ln{x})} dx} \\ = \int{\sin{(\ln{x})} d(\ln{x})} \\ = -\int{d\cos[(\ln{x})]} \\ = -\cos[(\ln{x})] + C$

$例25. 计算\int{\dfrac{x}{\sqrt{1 - x}} dx}$
$解:\\ 令t = \sqrt{1 - x}, x = 1 - t^2; dx = -2tdt\\ \int{\dfrac{x}{\sqrt{1 - x}} dx} \\ = \int{\dfrac{1 - t^2}{t} (-2tdt)} \\ = 2\int{(t^2 - 1)dt} \\ = \dfrac{2}{3}t^3 - 2t + C \\ = \dfrac{2}{3}(1 - x)\sqrt{1 - x} - 2\sqrt{1 - x} + C \\ = -\dfrac{2}{3}(2 + x)\sqrt{1 - x} + C$

$例26. 计算\int{\sqrt{a^2 - x^2}dx}$
$解:\\ 令x = a\sin{t}, t \in [-\dfrac{\pi}{2}, \dfrac{\pi}{2}] \\ t =\arcsin{\frac{x}{a}} \\ \int{\sqrt{a^2 - x^2}dx} \\ = \int{a\cos{t}d(a\sin{t})} \\ = a^2 \int{\cos^2{t} dt} \\ = a^2 \int{\dfrac{1 + \cos{2t}}{2} dt} \\ = a^2 \int{\dfrac{1 + \cos{2t}}{2} dt} \\ = \dfrac{a^2}{2} t + \dfrac{a^2}{4} \int{d\sin{2t}} \\ = \dfrac{a^2}{2} t + \dfrac{a^2}{4} \sin{2t} + C \\ = \dfrac{a^2}{2} t + \dfrac{a^2}{2} \sin{t}\cos{t} + C \\ =\dfrac{a^2}{2}\arcsin{\dfrac{x}{a}} + \dfrac{a^2}{2} \dfrac{x}{a} \dfrac{a^2 - x^2}{a} + C \\ =\dfrac{a^2}{2}\arcsin{\dfrac{x}{a}} + \dfrac{x\sqrt{a^2 - x^2}}{2} + C$

$例27. 计算\int{x \sin{x}dx}$
$解:\\ \int{x \sin{x}dx} \\ = \int{x d(-\cos{x})} \\ = -x\cos{x} - \int{(-\cos{x})dx} \\ = -x\cos{x} + \sin{x} + C$

$例28. 计算\int{\ln{x} dx}$
$解:\\ \int{\ln{x} dx} \\ = x \ln{x} - \int{x d{\ln{x}} } \\ = x \ln{x} - \int{x \dfrac{1}{x} dx } \\ = x \ln{x} - x + C$

$例29. 计算\int{ x\ln{x} dx}$
$解:\\ \int{ x\ln{x} dx} \\ = \int{\ln{x} d(\dfrac{1}{2}x^2)} \\ = \dfrac{1}{2}x^2 \ln{x} - \int{(\dfrac{1}{2}x^2) d(\ln{x})} \\ = \dfrac{1}{2}x^2 \ln{x} - \int{(\dfrac{1}{2}x) dx} \\ = \dfrac{1}{2}x^2 \ln{x} - \dfrac{1}{4}x^2 + C$

$例30.求\int{x^2e^{-x} dx}$
$解:\\ \int{x^2e^{-x} dx} \\ = \int{x^2 d(-e^{-x})} \\ = -x^2e^{-x} - \int{(-e^{-x})d(x^2)} \\ = -x^2e^{-x} + 2\int{xe^{-x}dx} \\ = -x^2e^{-x} - 2\int{xde^{-x}} \\ = -x^2e^{-x} - 2xe^{-x} + 2\int{e^{-x} dx} \\ = -x^2e^{-x} - 2xe^{-x} - 2e^{-x} + C \\ = -e^{-x}(x^2 + 2x + 2) + C$

$例31.求不定积分\int{e^{2x} \sin{x} dx}$
$解:\\ \int{e^{2x} \sin{x} dx} \\ = \int{e^{2x} d(-\cos{x}) } \\ = -e^{2x} \cos{x} - \int{(-\cos{x})de^{2x}} \\ = -e^{2x} \cos{x} + 2\int{\cos{x} e^{2x} dx} \\ = -e^{2x} \cos{x} + 2\int{e^{2x} d\sin{x}} \\ = -e^{2x} \cos{x} + 2e^{2x} \sin{x} - 2\int{\sin{x} de^{2x}} \\ = -e^{2x} \cos{x} + 2e^{2x} \sin{x} - 4\int{\sin{x} e^{2x}dx} \\ \int{e^{2x} \sin{x} dx} = \dfrac{e^{2x}}{5}(2 \sin{x} - \cos{x}) + C$

练习题
(一)单项选择题
$1.下列函数中,({\color{blue}{C} })是x\sin{x^2}的原函数.$
$A.-2\cos{x^2} \quad B.2\cos{x^2} \\ C.-\dfrac{1}{2}\cos{x^2} \quad D.\dfrac{1}{2}\cos{x^2}$
$解:\\ F(x) = \int{x\sin{x^2} dx} \\ = \dfrac{1}{2} \int{\sin{x^2} d(x^2)} \\ = -\dfrac{1}{2} \int{d(\cos{x^2})} \\ = -\dfrac{1}{2}\cos{x^2} + C$

$2.下列等式中,({\color{blue}{D} })是正确的.$
$A.d\int{f(x)dx} = f(x) \quad B.\dfrac{d}{dx}\int{f(x)dx} = f(x)dx \\ C.\dfrac{d}{dx}\int{f(x)dx} = f(x) + C \quad D.d\int{f(x)dx} = f(x)dx$

$3.若\int{f(x)e^{\frac{1}{x}}dx} = -e^{\frac{1}{x}} + C,则f(x) = ({\color{blue}{B} }).$
$A.\dfrac{1}{x} \quad B.\dfrac{1}{x^2} \quad C.-\dfrac{1}{x} \quad D.-\dfrac{1}{x^2}$
$解:\\ (-e^{\frac{1}{x}} + C)^{\prime} \\ =-e^{\frac{1}{x}}(x^{-1})^{\prime} \\ =\dfrac{1}{x^2}e^{\frac{1}{x}}$

$4.若f(x)满足\int{f(x)dx} = \sin{2x} + C,则f^{\prime}(x) = ({\color{blue}{C} }).$
$A.4\sin{2x} \quad B.2\cos{2x} \quad C.-4\sin{2x} \quad D.-2\cos{2x}$
$解:\\ f(x) = (\sin{2x} + C)^{\prime} = 2\cos{2x}\\ f^{\prime}(x) = (2\cos{2x})^{\prime} = -4\sin{2x}$

(二)填空题
$1.设\int{f(x)dx} = \dfrac{1}{1 + x^2} + C,则f(x) = \underline{\ \ {-\dfrac{2x}{(1 + x^2)^2}}\ \ }$
$解:\\ f(x) = (\dfrac{1}{1 + x^2} + C)^{\prime} \\ = -\dfrac{2x}{(1 + x^2)^2}$

$2.若函数F(x)与G(x)是同一个连续函数的原函数,\\ 则F(x)与G(x)之间有关系\underline{\ \ {\color{blue}{F(x) - G(x) = C(C为常数)} }\ \ }$
$解:\\ $

$3.若\int{f(x)dx} = e^{-x^2} + C,则f(x) = \underline{\ \ {\color{blue}{-2xe^{-x^2}} }\ \ }$
$解:\\ f(x) = (e^{-x^2} + C)^{\prime} = -2xe^{-x^2}$

$4.若\int{f(x)dx} = -\cos{x} + C,则f^{(n)}(x) = \underline{\ \ {\color{blue}{ \sin{(x + \dfrac{n}{2} \pi)} } }\ \ }$
$解:\\ f(x) = (-\cos{x} + C)^{\prime} = \sin{x} \\ f^{(1)}(x) = (\sin{x})^{\prime} = \cos{x} = \sin{(x + \dfrac{1}{2} \pi)} \\ f^{(2)}(x) = (\cos{x})^{\prime} = -\sin{x} = \sin{(x + \dfrac{2}{2} \pi)} \\ f^{(3)}(x) = (-\sin{x})^{\prime} = -\cos{x} = \sin{(x + \dfrac{3}{2} \pi)} \\ f^{(4)}(x) = (-\cos{x})^{\prime} = \sin{x} = \sin{(x + \dfrac{4}{2} \pi)} \\ \cdots \\ f^{(n)}(x) = \sin{(x + \dfrac{n}{2} \pi)}$

$5.\int{\dfrac{1}{1 - 2x}dx} = \underline{\ \ {\color{blue}{ -\dfrac{1}{2} \ln |1 - 2x| + C } }\ \ }$
$解:\\ \int{\dfrac{1}{1 - 2x}dx} \\ = -\dfrac{1}{2}\int{\dfrac{1}{1 - 2x}d(1 - 2x)} \\ = -\dfrac{1}{2} \ln |1 - 2x| + C$

$6.\int{\ln{x}d(\ln{x})} = \underline{\ \ {\color{blue}{ \dfrac{1}{2} (\ln{x})^2 + C} }\ \ }$
$解:\\ \int{\ln{x}d(\ln{x})} \\ = \dfrac{1}{2} \int{d(\ln{x})^2} \\ = \dfrac{1}{2} (\ln{x})^2 + C$

$7.\int{xf(x^2)f^{\prime}(x^2) dx} = \underline{\ \ {\color{blue}{\dfrac{1}{4} [f(x^2)]^2 + C } }\ \ }$
$解:\\ \int{xf(x^2)f^{\prime}(x^2) dx} \\ = \dfrac{1}{2} \int{f(x^2)f^{\prime}(x^2) dx^2} \\ = \dfrac{1}{2} \int{f(x^2) df(x^2)} \\ = \dfrac{1}{4} \int{d[f(x^2)]^2} \\ = \dfrac{1}{4} [f(x^2)]^2 + C$

$8.\int{f(x) dx} = x + C, 则\int{f(1 - x) dx} = \underline{\ \ {\color{blue}{x + C } }\ \ }$
$解:\\ f(x) = (x + C)^{\prime} = 1\\ \int{(1 - x)dx} = \int{dx} = x + C$

(三)解答题
$1.求2x + x^2的全体原函数.$
$解:\\ \int{(2x + x^2)dx} \\ = x^2 + \dfrac{1}{3}x^3 + C$

$2.求过点(0, 2)的曲线y = f(x), 使它在x点处的切线斜率为e^x.$
$解:\\ f^{\prime}(x) = e^x \\ f(x) = \int{e^x dx} = e^x + C \\ f(0) = 2 = e^0 + C \\ C = 1 \\ f(x) = e^x + 1$

$3.求不定积分\int{(1 - \sqrt{x})(\dfrac{1}{\sqrt{x}} + x) dx}.$
$解:\\ \int{(1 - \sqrt{x})(\dfrac{1}{\sqrt{x}} + x) dx} \\ = \int{[x^{-\frac{1}{2}} -1 + x -x^{\frac{3}{2}}]dx} \\ = 2\sqrt{x} -x + \dfrac{1}{2}x^2 - \dfrac{2}{5}x^{\frac{5}{2}} + C$

$4.求\int{\dfrac{e^{3 + x}}{2} dx}.$
$解:\\ \int{\dfrac{e^{3 + x}}{2} dx} \\ = \dfrac{1}{2}\int{e^{3 + x} d(3 + x)} \\ = \dfrac{1}{2}e^{3 + x} + C \\ = \dfrac{e^3}{2}e^x + C$

$5.求\int{\dfrac{2x^2 - \sqrt[3]{x}}{\sqrt{x}} dx}$
$解:\\ \int{\dfrac{2x^2 - \sqrt[3]{x}}{\sqrt{x}} dx} \\ = 2\int{x^{\frac{3}{2}} dx} - \int{ x^{-\frac{1}{6}} dx} \\ = \dfrac{4}{5}x^{\frac{5}{2}} - \dfrac{6}{5}x^{\frac{5}{6}} + C$

$6.求\int{e^{-2x} dx}$
$解:\\ \int{e^{-2x} dx} \\ = -\dfrac{1}{2} \int{e^{-2x} d(-2x)} \\ = -\dfrac{e^{-2x}}{2} + C$

$7.求\int{x^2\sqrt{x^3 + 1} dx}$
$解:\\ \int{x^2\sqrt{x^3 + 1} dx} \\ = \dfrac{1}{3} \int{\sqrt{x^3 + 1} d(x^3 + 1)} \\ = \dfrac{2}{9} \int{d(x^3 + 1)^{\frac{3}{2}}} \\ = \dfrac{2}{9}(x^3 + 1)^{\frac{3}{2}} + C$

$8.求\int{\dfrac{1}{x(1 + 3\ln{x})} dx}$
$解:\\ \int{\dfrac{1}{x(1 + 3\ln{x})} dx} \\ = \int{\dfrac{1}{(1 + 3\ln{x})} d\ln{x}} \\ = \dfrac{1}{3}\int{\dfrac{1}{(1 + 3\ln{x})} d(1 + 3\ln{x})} \\ = \dfrac{\ln|1 + 3\ln{x}|}{3} + C$

$9.求\int{ \dfrac{\ln{x}}{x} dx}$
$解:\\ \int{ \dfrac{\ln{x}}{x} dx} \\ = \int{ \ln{x} d\ln{x}} \\ = \dfrac{(\ln{x})^2}{2} + C$

$10.求\int{\dfrac{\cos{\dfrac{1}{x}}}{x^2} dx}$
$解:\\ \int{\dfrac{\cos{\dfrac{1}{x}}}{x^2} dx} \\ = -\int{\cos{\dfrac{1}{x}} d\dfrac{1}{x}} \\ = -\int{ d\sin{\dfrac{1}{x}}} \\ = -\sin{\dfrac{1}{x}} + C$

$11.求\int{\dfrac{1}{\sqrt{x}(1 + x)} dx}$
$解:\\ \int{\dfrac{1}{\sqrt{x}(1 + x)} dx} \\ = \int{\dfrac{1}{\sqrt{x}[1 + (\sqrt{x})^2]} 2\sqrt{x}d\sqrt{x}} \\ = 2\int{\dfrac{1}{[1 + (\sqrt{x})^2]}d\sqrt{x}} \\ = 2\arctan{\sqrt{x}} + C$

$12.求\int{\dfrac{x + (\arctan{x})^2}{1 + x^2} dx}$
$解:\\ \int{\dfrac{x + (\arctan{x})^2}{1 + x^2} dx} \\ = \int{\dfrac{x}{1 + x^2} dx} + \int{\dfrac{(\arctan{x})^2}{1 + x^2} dx} \\ = \dfrac{1}{2}\int{\dfrac{1}{1 + x^2} d(1 + x^2)} + \int{(\arctan{x})^2 d\arctan{x}} \\ = \dfrac{\ln(1 + x^2)}{2} + \dfrac{\arctan^3{x}}{3} + C$

$13.求\int{x\cos{x} dx}$
$解:\\ \int{x\cos{x} dx} \\ = \int{xd\sin{x}} \\ = x\sin{x} - \int{\sin{x}dx} \\ = x\sin{x} + \cos{x} + C$

$14.求\int{\dfrac{\ln{x}}{x^2} dx}$
$解:\\ \int{\dfrac{\ln{x}}{x^2} dx} \\ = -\int{\ln{x} dx^{-1}} \\ = -\dfrac{\ln{x}}{x} + \int{\dfrac{1}{x} d\ln{x}} \\ = -\dfrac{\ln{x}}{x} + \int{\dfrac{1}{x^2} dx} \\ = -\dfrac{\ln{x}}{x} - \dfrac{1}{x} + C$

$15.求\int{\cos{\sqrt{x + 1}} dx}$
$解:\\ 令t = \sqrt{x + 1}, x = t^2 -1,dx = 2tdt\\ \int{\cos{\sqrt{x + 1}} dx} \\ = \int{\cos{t} 2tdt} \\ = 2\int{t d\sin{t}} \\ = 2t\sin{t} - 2\int{\sin{t}dt} \\ = 2t\sin{t} + 2\int{d\cos{t}} \\ = 2t\sin{t} + 2\cos{t} + C \\ = 2\sqrt{x + 1}\sin{\sqrt{x + 1}} + 2\cos{\sqrt{x + 1}} + C$

$16.求\int{e^{2x}\cos{e^x} dx}$
$解:\\ \int{e^{2x}\cos{e^x} dx} \\ = \int{e^{x}\cos{e^x} de^x} \\ = \int{e^{x}d\sin{e^x}} \\ = e^x\sin{e^x} - \int{\sin{e^x} de^x} \\ = e^x\sin{e^x} + \int{ d\cos{e^x}} \\ = e^x\sin{e^x} + \cos{e^x} + C$