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Question

$\int\limits_{\frac{1}{\sqrt[3]{3}}}^{1} \frac{dx}{(x^{3}+\frac{1}{3})\times\sqrt[3]{x^{3}+\frac{1}{3}} }$

bathoainguyen · Answer 1 · 6/14/2014 11:53:44 AM

$I=\int_{\sqrt[3]{3}}^{1}\frac{dx}{x^4\sqrt[3]{(1+\frac{1}{3x^3})^4}}\\

t=\sqrt[3]{1+\frac{1}{3x^3}}\Rightarrow t^3=1+\frac{1}{3x^3}\Rightarrow 3t^2dt=-\frac{dx}{x^4}\\

I=\int_{\sqrt[3]{\frac{4}{3}}}^{\sqrt[3]{\frac{10}{9}}}\frac{3t^2dt}{t^4}==\int_{\sqrt[3]{\frac{4}{3}}}^{\sqrt[3]{\frac{10}{9}}}\frac{3dt}{t^2}=-\frac{3}{t}\\

I=-3(\sqrt[3]{\frac{9}{10}}-\sqrt[3]{\frac{3}{4}})$

Trả lời 14-06-14 11:53 AM

bathoainguyen
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sal.yglover · Answer 2 · 5/31/2014 12:25:19 PM

= $\int\limits_{\frac{1}{\sqrt[3]{3}}}^{1}\frac{(x^{2}\times x )dx}{x^{3}\times (x^{3}+\frac{1}{3})\times \sqrt[3]{x^{3}+\frac{1}{3}}}$
= $\int\limits_{\frac{1}{\sqrt[3]{3}}}^{1}\frac{x^{2}dx}{x^{3}\times (x^{3}+\frac{1}{3})\times \sqrt{x^{3}+\frac{1}{3}}}$
Đặt $x=\frac{1}{t}$ $\Rightarrow dx=\frac{-dt}{x^{2}}$
Khi $x=\frac{1}{\sqrt[3]{3}} \Rightarrow t=\sqrt[3]{3}$

Khi $x=1 \Rightarrow t=1$

$\Rightarrow I=\int\limits_{\sqrt[3]{3}}^{1}\frac{\frac{-dt}{t^{2}}}{(\frac{1}{t^{3}}+\frac{1}{3})\times \sqrt[3]{\frac{1}{t^{3}}+\frac{1}{3}}} $
=$\int\limits_{1}^{\sqrt[3]{3}}\frac{\frac{dt}{t^{2}}}{\frac{3+t^{3}}{3t^{3}}\times \frac{\sqrt[3]{3+t^{3}}}{\sqrt[3]{3}t}}$
=$3\sqrt[3]{3}\times \int\limits_{1}^{\sqrt[3]{3}}\frac{t^{2}dt}{(3+t^{3})\times \sqrt[3]{3+t^{3}}}$
Đặt $u=\sqrt[3]{3+t^{3}} \Rightarrow u^{3}=t^{3}+3\Rightarrow u^{2}du=t^{2}dt$
Khi $t=1\Rightarrow u=\sqrt[3]{4}$

Khi $t=\sqrt[3]{3}\Rightarrow u=\sqrt[3]{6}$
$\Rightarrow I=3\sqrt[3]{3}\int\limits_{\sqrt[3]{4}}^{\sqrt[3]{6}}\frac{u^{2}du}{u^{4}}$

$=3\sqrt[3]{3}\times (\frac{1}{\sqrt[3]{4}}-\frac{1}{\sqrt[3]{6}})$