Nguyen Kieu Phuong

$A=yz_{x}^{^{'}}-xz_{y}^{'};z=r^{3}+3^{r}+ln(1+2r),r=\sqrt{x^{2}+y^{2}}$

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$=\lim_{x\to0}\frac{cotx.(-sin^2x)}{x}=-\lim_{x\to0}\frac{cosx.sinx}{x}=-1$

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A= $\int_{0}^{+\propto}\frac{x^{2}}{\sqrt{1+x^{2}}}dx$

$=\lim_{b\to+\infty }\int_{0}^{b}\frac{x^2}{\sqrt{1+x^2}}dx$

$\int \frac{x^2}{\sqrt{1+x^2}}dx=\int \frac{x^2+1-1}{\sqrt{1+x^2}}dx=\int (\sqrt{x^2+1}-\frac{1}{\sqrt{x^2+1}})dx$

$I_1=\int \frac{1}{\sqrt{x^2+1}}dx=ln|x+\sqrt{x^2+1}|+c$

$I_2=\int \sqrt{x^2+1}dx=\int \sqrt{tan^2x+1}dx=\int \frac{1}{cosx}dx$

đặt $t=tan(\frac{x}{2})=>dx=\frac{2dt}{t^2+1};cosx=\frac{1-t^2}{1+t^2}$

$=>I_2=\int \frac{2dt}{1-t^2}=ln|1-t^2|+c$

Tìm A = $\int\frac{\sqrt{x+1}}{x}dx$

đặt $t=\sqrt{x+1}=>t^2=x+1=>2tdt=dx$

$=>A=\int \frac{2tdt}{t^2-1}=\int \frac{t-1+t+1}{(t-1)(t+1)}dt=ln|t^2-1|$

$=\lim_{x\to0}(\frac{tan^3x-x^3}{x^3.tan^3x})$

$=\lim_{x\to0}\frac{\frac{x^3}{cos^3x}-x^3}{x^3.tan^3x}$

$=\lim_{x\to0}\frac{\frac{1}{cos^3x}-1}{\frac{x^3}{cos^3x}}$

$=\lim_{x\to0}\frac{1-cos^3x}{x^3}$

$=\lim_{x\to0}\frac{\frac{x^2}{2}(1+cosx+cos^2)}{x^3}$

$=\lim_{x\to0}\frac{(1+cosx+cos^2x)}{2x}=+\infty$