错题整理

1.$\int_{1}^{+\infty}\frac{arctanx}{x^2(1+x^2)}\,dx=?$
解
令$x=tanu$

$$原式=\int_{\frac{\pi}{4}}^{\frac{\pi}{2}}\frac{u}{tan^2usec^2u}\,d(tanu)$$ $$=\int_{\frac{\pi}{4}}^{\frac{\pi}{2}}ucot^2u\,du$$ $$=\int_{\frac{\pi}{4}}^{\frac{\pi}{2}}u(csc^2u-1)du$$ $$=-\int_{\frac{\pi}{4}}^{\frac{\pi}{2}}u\,d(cotu)-\int_{\frac{\pi}{4}}^{\frac{\pi}{2}}u\,du$$ $$=-(ucotu\bigg|_{\frac{\pi}{4}}^{\frac{\pi}{2}}-\int_{\frac{\pi}{4}}^{\frac{\pi}{2}}cotu\,du)-\frac{1}{2}u^2\bigg|_{\frac{\pi}{4}}^{\frac{\pi}{2}}$$ $$=-ucotu\bigg|_{\frac{\pi}{4}}^{\frac{\pi}{2}}+\int_{\frac{\pi}{4}}^{\frac{\pi}{2}}\frac{1}{sinu}\,d(sinu)-\frac{1}{2}u^2\bigg|_{\frac{\pi}{4}}^{\frac{\pi}{2}}$$ $$=-ucotu\bigg|_{\frac{\pi}{4}}^{\frac{\pi}{2}}+ln(sinx)\bigg|_{\frac{\pi}{4}}^{\frac{\pi}{2}}-\frac{1}{2}u^2\bigg|_{\frac{\pi}{4}}^{\frac{\pi}{2}}$$ $$=\frac{\pi}{4}+\frac{ln2}{2}-\frac{3}{32}\pi^2$$

2.$\int\frac{\sqrt[3]{x+2}-2\sqrt[4]{x+2}}{(\sqrt[4]{x+2}-\sqrt[6]{x+2})^7}\,dx$
解
取根号的最小公倍数
令$t=\sqrt[12]{x+2}$

$$原式=\int\frac{t^4-2t^3}{(t^3-t^2)^7}\,d(t^{12}-2)$$ $$原式=\int\frac{12t^{15}-24t^{14}}{(t^3-t^2)^7}\,dt$$ $$=\int\frac{12t-24}{(t-1)^7}\,dt$$ $$=12\int\frac{1}{(t-1)^6}\,dt-12\int\frac{1}{(t-1)^7}\,dt$$ $$=\frac{2}{(\sqrt[12]{x+2}-1)^6}-\frac{12}{5(\sqrt[12]{x+2}-1)^5}+C$$

3.$\int\frac{1}{\sqrt[5]{x^3(x+1)^7}}\,\textrm{d}x$

4.求f(x)，使对任意实数x有如下等式成立：$f(2+x)+2f(1-x)=x^2$

5.$\lim_{x \to 0}\frac{x^3(\frac{1}{x^2}+sin\frac{1}{x^2})}{x+sinx}$

6.$若x=1是函数f(x)=\frac{4\sqrt{x+3}-ax-b}{(x-c)^2}的可去间断点，则a=,b=,c=$

7.$若f(x)在x=x_0点处连续，试证明：函数|f(x)|在x=x_0处也连续。$

8.$设f(x)=x(x+1)(x+2)......(x+2n),则f'(-n)=$
解

$$f(x)=(x+n)x(x+1)(x+2)......(x+2n)$$ $$令h(x)=x+n, g(x)=x(x+1)......(x+2n)$$ $$f'(x)=g(x)+h(x)g'(x)$$ $$f'(-n)=g(-n)\\=-n(-n+1)...(-n+n-2)(-n+n-1)(-n+n+1)(-n+n+2)...(-n+2n)\\=(-1)^n(n!)^2$$

9.$计算定积分\int_{\frac{1}{2}}^{2}(1+x-\frac{1}{x})e^{x+\frac{1}{x}}\,dx$

10.$设函数f(x)在[0,1]上连续，在(0,1)内二阶可导，且f''(x)\geq0，请证明：\int_{0}^{1}f(x^\alpha)\,dx\geq f(\frac{1}{\alpha+1})(\alpha>0)$

11.$函数f(x)=\frac{1}{2+x}带拉格朗日余项的二阶麦克劳林公式为f(x)=$

12.$函数f(x)=xe^{1-x}在x_0=1处带拉格朗日余项的三阶泰勒公式为f(x)=$

13.$设f(x)=\int_{0}^{1}|t(t-x)|\,dt,则f'(x)=$

14.$\int_{-\pi/4}^{\pi/4} 5cosx\cdot arctane^x\,dx$

15.$求摆线\begin{cases}{x=a(t-sint)}\\y=a(1-cost)\end{cases}(0\leq t\leq 2\pi)关于x旋转一周形成的旋转体体积$

16.$若x=1是函数f(x)=\frac{4\sqrt{x+3}-ax-b}{(x-c)^2}的可去间断点，则a=,b=,c=.$

17.$\lim_{x\to+\infty}(\frac{a_1^x+a_2^x+...+a_n^x}{n})^{\frac{1}{x}}=$

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