K=$\int\limits_{1}^{4}$$\frac{dx}{x^{2}+x\sqrt{x}}$
Trả lời 04-03-20 09:01 AM
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Câu 1 : cho tích phân $\int\limits_{e}^{e^{3}}(\frac{1}{(lnx)^{2}}-\frac{1}{ln(x)}).dx=ae^{3}+be.$ Tính a+b =?
Trả lời 28-02-17 10:15 AM
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$I=\int\limits_{1}^{4}e^{\sqrt{x}}dx$Giúp mình với nha! Cảm ơn các bạn.
Trả lời 24-02-16 11:36 AM
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Bỏ dấu ngoặc rồi tính:
$a) (27 + 65) + (346 - 27 - 65).$ $b) (42 - 69 + 17) - (42 + 17)$.
Trả lời 27-01-16 08:17 PM
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Bỏ dấu ngoặc rồi tính:
$a) (27 + 65) + (346 - 27 - 65).$ $b) (42 - 69 + 17) - (42 + 17)$.
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I1 = $\int\limits_{0}^{\frac{\pi }{2}}(2x-1)cos^{2}xdx$I2 =$\int\limits_{1}^{3}\frac{1}{(x+3)^3}lnxdx$I3 = $\int\limits_{0}^{\frac{\pi }{4}}\frac{x}{1+cos2x}dx$
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I1 = $\int\limits_{0}^{\frac{\pi }{2}}(2x-1)cos^{2}xdx$I2 =$\int\limits_{1}^{3}\frac{1}{(x+3)^3}lnxdx$I3 = $\int\limits_{0}^{\frac{\pi }{4}}\frac{x}{1+cos2x}dx$
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I1 = $\int\limits_{0}^{\frac{\pi }{2}}(2x-1)cos^{2}xdx$I2 =$\int\limits_{1}^{3}\frac{1}{(x+3)^3}lnxdx$I3 = $\int\limits_{0}^{\frac{\pi }{4}}\frac{x}{1+cos2x}dx$
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K=$\int\limits_{1}^{4}$$\frac{dx}{x^{2}+x\sqrt{x}}$
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K=$\int\limits_{1}^{4}$$\frac{dx}{x^{2}+x\sqrt{x}}$
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$\int\limits_{1}^{2} \frac{dx}{\sqrt{x+1} + \sqrt{x-1}}$$\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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