Cho $A=\int\limits_{2}^{3}\frac{x^2}{(x+1)^3+(x+1)^5}\mathrm{d}x=-\frac{41}{288}-2\cdot \arctan a+2\cdot \arctan b$. Tính $S=a+b$ ?A. $S=4$B. $S=-4$C. $S=7$D. $S=0$
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K=$\int\limits_{1}^{4}$$\frac{dx}{x^{2}+x\sqrt{x}}$
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Cho $\int\limits_{2}^{3} \frac{1}{x^3+x^5}dx=\frac{a}{72}+\frac{3}{2}\ln(b)-\ln(c)$. Tổng $a+b+c$ bằng?A. 10B. -5C. 4D. -10
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Biết $\int\limits_{1}^{2}\frac{dx}{\sqrt{x+1}.\sqrt{x+2}}$=$\ln (a\sqrt{6}+b\sqrt{3}+c\sqrt{2}+d)$ với a,b,c,d là các số nguyên.Tính P=a+b+c+d
Trả lời 29-05-18 11:59 AM
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Giả sử: $I=\int\limits_{0}^{b\frac{\Pi }{4}}\sin 3x\sin 2xdx=a+b\frac{\sqrt{2}}{2}$, khi đó, giá trị a+b=?A, -1/6B, 3/10C, -1/10D, 1/5
Trả lời 19-01-18 02:16 PM
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Cho tích phân : $\int\limits_{1}^{2}\frac{1-x}{x^{2}}.e^{x}.dx=ae^{2}+be\Rightarrow a+b=?$Thanks trước nhé !!!
Trả lời 28-02-17 10:01 AM
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TÍNH TÍCH PHÂN $I=\int\limits_{1}^{2}\frac{X^2-1}{X^4+1}$
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$\int\limits_{0}^{3}\frac{2x-1}{x^{2}+4x+5}$
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$I=\int\limits_{0}^{1} \frac{\sqrt{x-1}}{\sqrt{(x+1)^5}} dx$
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tính tích phân $I=\int\limits_{0}^{1}\frac{x^{3}\times dx}{x+\sqrt{x^{2}+1}}$
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K=$\int\limits_{1}^{4}$$\frac{dx}{x^{2}+x\sqrt{x}}$
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K=$\int\limits_{1}^{4}$$\frac{dx}{x^{2}+x\sqrt{x}}$
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$\int\limits_{1}^{\sqrt{3}}\frac{\sqrt{9+3x^{2}}}{x^{2}}dx$
Trả lời 08-01-15 05:07 PM
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$\int\limits_{1}^{\sqrt{3}}\frac{\sqrt{9+3x^{2}}}{x^{2}}dx$
Trả lời 07-01-15 03:50 PM
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$1. \int\limits_{0}^{1} \frac{x^{2}}{x^{2}+1}$ dx$2 \int\limits_{0}^{1} \frac{x}{x^{4}+1}$ dx$3. \int\limits_{0}^{\frac{1}{3}} \frac{1}{4x^{2}-1} dx$
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$\int\limits_{0}^{2}x.e^{-x^{2}}$
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$\int\limits_{0}^{1}$$\frac{dx}{\sqrt{x^{2}+1}}$
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$1)I=\int\limits_{}^{} \frac{2-\sqrt{4-x^2}}{3x^4}dx$$2) I=\int\limits_{}^{} \frac{1}{x\sqrt{4-x^2}}dx$$3) I=\int\limits_{}^{}\frac{1}{x\sqrt{4-\ln ^2\left| {x} \right|}}dx$$4)I=\int\limits_{}^{}\frac{x+sinx}{1+cosx}dx$
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$\int\limits_{1}^{2}\frac{1-x^{2}}{x+x^{3}}dx$
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$I=\int\limits_{1}^{3}\frac{1+x(2lnx-1)}{x(x+1)^{2}}dx$
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