K=$\int\limits_{1}^{4}$$\frac{dx}{x^{2}+x\sqrt{x}}$
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cho a+2b-3c=0;bc+2ac-3ab=o chứng minh a=b=c
Trả lời 26-10-17 04:19 AM
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cho a+2b-3c=0;bc+2ac-3ab=o chứng minh a=b=c
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cho a+2b-3c=0;bc+2ac-3ab=o chứng minh a=b=c
Trả lời 25-10-17 07:46 AM
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Nguyên hàm của x/( (căn của x^2 -1)+x)
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K=$\int\limits_{1}^{4}$$\frac{dx}{x^{2}+x\sqrt{x}}$
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K=$\int\limits_{1}^{4}$$\frac{dx}{x^{2}+x\sqrt{x}}$
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$\int\limits_{\frac{-\pi }{2}}^{0}\frac{\sin 2x}{(2+\sin x)^2} dx$
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$\int\limits_{1}^{2} \frac{dx}{\sqrt{x+1} + \sqrt{x-1}}$$\int\limits_{\frac{\pi}{6}}^{\frac{\pi}{4}} \frac{3-sin^{2}x}{\sin^{2}x} dx$$\int\limits_{0}^{\frac{\pi }{2}} (tan2x+cos2x)^{2}dx$$\int\limits_{0}^{\frac{\pi}{4}} \frac{x+sin2x}{cos^{2}x}dx$
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$\int\limits_{1}^{2} \frac{dx}{\sqrt{x+1} + \sqrt{x-1}}$$\int\limits_{\frac{\pi}{6}}^{\frac{\pi}{4}} \frac{3-sin^{2}x}{\sin^{2}x} dx$$\int\limits_{0}^{\frac{\pi }{2}} (tan2x+cos2x)^{2}dx$$\int\limits_{0}^{\frac{\pi}{4}} \frac{x+sin2x}{cos^{2}x}dx$
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$\int\limits_{1}^{2} \frac{dx}{\sqrt{x+1} + \sqrt{x-1}}$$\int\limits_{\frac{\pi}{6}}^{\frac{\pi}{4}} \frac{3-sin^{2}x}{\sin^{2}x} dx$$\int\limits_{0}^{\frac{\pi }{2}} (tan2x+cos2x)^{2}dx$$\int\limits_{0}^{\frac{\pi}{4}} \frac{x+sin2x}{cos^{2}x}dx$
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$\int\limits_{1}^{\sqrt{3}}\frac{\sqrt{9+3x^{2}}}{x^{2}}dx$
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$\int\limits_{1}^{\sqrt{3}}\frac{\sqrt{9+3x^{2}}}{x^{2}}dx$
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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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1. $\int\limits_{1}^{2} \frac{\sqrt{1+x^2}}{x^4}dx$2. $\int\limits_{0}^{1}\frac{e^x(1+x)}{1+xe^x}dx$3. $\int\limits_{0}^{\pi }\frac{xsinx}{1+cos^2 x}dx$4. $\int\limits_{0}^{ln2}\sqrt{e^x-1}dx$
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$\int\limits_{0}^{\frac{\pi }{4}}x\tan x.dx$
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$\int\limits_{0}^{\frac{\pi}{4} } \sqrt{\sin x} dx $
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tính tích phân$ \int\limits_{1}^{2}\frac{1-x^{2}}{1+x^{4}}$dx
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tính tích phân$ \int\limits_{1}^{2}\frac{1-x^{2}}{1+x^{4}}$dx
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$\int_1^2 \dfrac{\sqrt{x+1}}{ x+\sqrt{x^2-1}}dx$
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