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Tính tích phân: $arc$ $\cos x/\sqrt{(1-x^2)^3} $
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$4$ $+$ $2$ $+$ $3$ $=$ $5$$8$ $+$ $4$ $+$ $6$ $=$ $10$$3$ $+$ $2$ $+$ $1$ $=$ $2$$6$ $+$ $4$ $+$ $2$ $=$ $?$
Trả lời 06-05-16 07:49 PM
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$4$ $+$ $2$ $+$ $3$ $=$ $5$$8$ $+$ $4$ $+$ $6$ $=$ $10$$3$ $+$ $2$ $+$ $1$ $=$ $2$$6$ $+$ $4$ $+$ $2$ $=$ $?$
Trả lời 06-05-16 07:44 PM
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$I=\int\limits_{0}^{1} \frac{\sqrt{x-1}}{\sqrt{(x+1)^5}} dx$
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$\int\limits_{\frac{\pi }{6}}^{\frac{\pi }{3}}\sqrt{tan^2x+cot^2x-2}dx$
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$\int\limits_{\frac{\pi }{6}}^{\frac{\pi }{3}}\sqrt{tan^2x+cot^2x-2}dx$
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$1. \int\limits_{0}^{1} \frac{x^{2}}{x^{2}+1}$ dx$2 \int\limits_{0}^{1} \frac{x}{x^{4}+1}$ dx$3. \int\limits_{0}^{\frac{1}{3}} \frac{1}{4x^{2}-1} dx$
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$I=\int\limits_{1}^{3}\frac{1+x(2lnx-1)}{x(x+1)^{2}}dx$
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