Quay lại đáp án

	3	thêm 7 ký tự trong đáp án
$I=\int\limits_{0}^{1}\left ( \frac{1}{x^4+x^2+1}+x^3(1-x^2)^3 \right )dx$$=\int\limits_{0}^{1}\frac{1}{x^4+x^2+1}dx+\int\limits_{0}^{1}x^3(1-x^2)^3dx $ $=I_1+I_2$ Trong đó$I_2=\int\limits_{0}^{1}x^3(1-x^2)^3dx=\left[ {-\frac{x^{10}}{10}+\frac{3x^{8}}{10}-\frac{x^{6}}{2}+\frac{x^{4}}{4}} \right]_0^1=\frac{1}{40}$$I_1=\int\limits_{0}^{1}\frac{1}{x^4+x^2+1}dx=\frac{1}{2}\int\limits_{0}^{1}\left ( \frac{x+1}{x^2+x+1}-\frac{x-1}{x^2-x+1}\right )dx$ $=\frac{1}{4}\int\limits_{0}^{1}\left ( \frac{d(x^2+x+1)}{x^2+x+1}-\frac{d(x^2-x+1)}{x^2-x+1}\right )+\frac{1}{4}\int\limits_{0}^{1}\left ( \frac{1}{x^2+x+1}+\frac{1}{x^2-x+1}\right )dx$ $=\left[ {\frac{1}{4}\ln\left\| {\frac{x^2+x+1}{x^2-x+1}}\right\|}+2\sqrt 3\arctan\frac{x-1}{\sqrt 3} +2\sqrt 3\arctan\frac{x-1}{\sqrt 3}\right]_0^1=\frac{\pi \sqrt 3}{12}+\frac{\ln 2}{4}$ $I=\int\limits_{0}^{1}\left ( \frac{1}{x^4+x^2+1}+x^3(1-x^2)^3 \right )dx$$=\int\limits_{0}^{1}\frac{1}{x^4+x^2+1}dx+\int\limits_{0}^{1}x^3(1-x^2)^3dx $ $=I_1+I_2$ Trong đó$I_2=\int\limits_{0}^{1}x^3(1-x^2)^3dx=\left[ {-\frac{x^{10}}{10}+\frac{3x^{8}}{10}-\frac{x^{6}}{2}+\frac{x^{4}}{4}} \right]_0^1=\frac{1}{40}$$I_1=\int\limits_{0}^{1}\frac{1}{x^4+x^2+1}dx=\frac{1}{2}\int\limits_{0}^{1}\left ( \frac{x+1}{x^2+x+1}-\frac{x-1}{x^2-x+1}\right )dx$$=\frac{1}{4}\int\limits_{0}^{1}\left ( \frac{d(x^2+x+1)}{x^2+x+1}-\frac{d(x^2-x+1)}{x^2-x+1}\right )+\frac{1}{4}\int\limits_{0}^{1}\left ( \frac{1}{x^2+x+1}+\frac{1}{x^2-x+1}\right )dx=$ $\left[ {\frac{1}{4}\ln\left\| {\frac{x^2+x+1}{x^2-x+1}}\right\|}+2\sqrt 3\arctan\frac{x-1}{\sqrt 3} +2\sqrt 3\arctan\frac{x-1}{\sqrt 3}\right]_0^1=\frac{\pi \sqrt 3}{12}+\frac{\ln 2}{4}$ $I=\int\limits_{0}^{1}\left ( \frac{1}{x^4+x^2+1}+x^3(1-x^2)^3 \right )dx$$=\int\limits_{0}^{1}\frac{1}{x^4+x^2+1}dx+\int\limits_{0}^{1}x^3(1-x^2)^3dx $ $=I_1+I_2$ Trong đó$I_2=\int\limits_{0}^{1}x^3(1-x^2)^3dx=\left[ {-\frac{x^{10}}{10}+\frac{3x^{8}}{10}-\frac{x^{6}}{2}+\frac{x^{4}}{4}} \right]_0^1=\frac{1}{40}$$I_1=\int\limits_{0}^{1}\frac{1}{x^4+x^2+1}dx=\frac{1}{2}\int\limits_{0}^{1}\left ( \frac{x+1}{x^2+x+1}-\frac{x-1}{x^2-x+1}\right )dx$ $=\frac{1}{4}\int\limits_{0}^{1}\left ( \frac{d(x^2+x+1)}{x^2+x+1}-\frac{d(x^2-x+1)}{x^2-x+1}\right )+\frac{1}{4}\int\limits_{0}^{1}\left ( \frac{1}{x^2+x+1}+\frac{1}{x^2-x+1}\right )dx$ $=\left[ {\frac{1}{4}\ln\left\| {\frac{x^2+x+1}{x^2-x+1}}\right\|}+2\sqrt 3\arctan\frac{x-1}{\sqrt 3} +2\sqrt 3\arctan\frac{x-1}{\sqrt 3}\right]_0^1=\frac{\pi \sqrt 3}{12}+\frac{\ln 2}{4}$

	2	thêm 40 ký tự trong đáp án		Sửa 20-10-12 12:05 AM Lee 1
$I=\int\limits_{0}^{1}\left ( \frac{1}{x^4+x^2+1}+x^3(1-x^2)^3 \right )dx$$=\int\limits_{0}^{1}\frac{1}{x^4+x^2+1}dx+\int\limits_{0}^{1}x^3(1-x^2)^3dx $ $=I_1+I_2$ Trong đó$I_2=\int\limits_{0}^{1}x^3(1-x^2)^3dx=\left[ {-\frac{x^{10}}{10}+\frac{3x^{8}}{10}-\frac{x^{6}}{2}+\frac{x^{4}}{4}} \right]_0^1=\frac{1}{40}$$I_1=\int\limits_{0}^{1}\frac{1}{x^4+x^2+1}dx=\frac{1}{2}\int\limits_{0}^{1}\left ( \frac{x+1}{x^2+x+1}-\frac{x-1}{x^2-x+1}\right )dx$$=\frac{1}{4}\int\limits_{0}^{1}\left ( \frac{d(x^2+x+1)}{x^2+x+1}-\frac{d(x^2-x+1)}{x^2-x+1}\right )+\frac{1}{4}\int\limits_{0}^{1}\left ( \frac{1}{x^2+x+1}+\frac{1}{x^2-x+1}\right )dx=$ $\left[ {\frac{1}{4}\ln\left\| {\frac{x^2+x+1}{x^2-x+1}}\right\|}+2\sqrt 3\arctan\frac{x-1}{\sqrt 3} +2\sqrt 3\arctan\frac{x-1}{\sqrt 3}\right]_0^1=\frac{\pi \sqrt 3}{12}+\frac{\ln 2}{4}$ $I=\int\limits_{0}^{1}\left ( \frac{1}{x^4+x^2+1}+x^3(1-x^2)^3 \right )dx$$=\int\limits_{0}^{1}\frac{1}{x^4+x^2+1}dx+\int\limits_{0}^{1}x^3(1-x^2)^3dx $ $=I_1+I_2$ Trong đó$I_2=\int\limits_{0}^{1}x^3(1-x^2)^3dx=\left[ {-\frac{x^{10}}{10}+\frac{3x^{8}}{10}-\frac{x^{6}}{2}+\frac{x^{4}}{4}} \right]_0^1=\frac{1}{40}$$I_1=\int\limits_{0}^{1}\frac{1}{x^4+x^2+1}dx=\frac{1}{2}\int\limits_{0}^{1}\left ( \frac{x+1}{x^2+x+1}-\frac{x-1}{x^2-x+1}\right )dx$$=\frac{1}{4}\int\limits_{0}^{1}\left ( \frac{d(x^2+x+1)}{x^2+x+1}-\frac{d(x^2-x+1)}{x^2-x+1}\right )+\frac{1}{4}\int\limits_{0}^{1}\left ( \frac{1}{x^2+x+1}+\frac{1}{x^2-x+1}\right )dx$ $\left[ {\frac{1}{4}\ln\left\| {\frac{x^2+x+1}{x^2-x+1}}\right\|}+2\sqrt 3\arctan\frac{x-1}{\sqrt 3} +2\sqrt 3\arctan\frac{x-1}{\sqrt 3}\right]_0^1$ $I=\int\limits_{0}^{1}\left ( \frac{1}{x^4+x^2+1}+x^3(1-x^2)^3 \right )dx$$=\int\limits_{0}^{1}\frac{1}{x^4+x^2+1}dx+\int\limits_{0}^{1}x^3(1-x^2)^3dx $ $=I_1+I_2$ Trong đó$I_2=\int\limits_{0}^{1}x^3(1-x^2)^3dx=\left[ {-\frac{x^{10}}{10}+\frac{3x^{8}}{10}-\frac{x^{6}}{2}+\frac{x^{4}}{4}} \right]_0^1=\frac{1}{40}$$I_1=\int\limits_{0}^{1}\frac{1}{x^4+x^2+1}dx=\frac{1}{2}\int\limits_{0}^{1}\left ( \frac{x+1}{x^2+x+1}-\frac{x-1}{x^2-x+1}\right )dx$$=\frac{1}{4}\int\limits_{0}^{1}\left ( \frac{d(x^2+x+1)}{x^2+x+1}-\frac{d(x^2-x+1)}{x^2-x+1}\right )+\frac{1}{4}\int\limits_{0}^{1}\left ( \frac{1}{x^2+x+1}+\frac{1}{x^2-x+1}\right )dx=$ $\left[ {\frac{1}{4}\ln\left\| {\frac{x^2+x+1}{x^2-x+1}}\right\|}+2\sqrt 3\arctan\frac{x-1}{\sqrt 3} +2\sqrt 3\arctan\frac{x-1}{\sqrt 3}\right]_0^1=\frac{\pi \sqrt 3}{12}+\frac{\ln 2}{4}$

	1	tạo mới		Trả lời 20-10-12 12:03 AM Lee 1
$I=\int\limits_{0}^{1}\left ( \frac{1}{x^4+x^2+1}+x^3(1-x^2)^3 \right )dx$$=\int\limits_{0}^{1}\frac{1}{x^4+x^2+1}dx+\int\limits_{0}^{1}x^3(1-x^2)^3dx $ $=I_1+I_2$ Trong đó$I_2=\int\limits_{0}^{1}x^3(1-x^2)^3dx=\left[ {-\frac{x^{10}}{10}+\frac{3x^{8}}{10}-\frac{x^{6}}{2}+\frac{x^{4}}{4}} \right]_0^1=\frac{1}{40}$$I_1=\int\limits_{0}^{1}\frac{1}{x^4+x^2+1}dx=\frac{1}{2}\int\limits_{0}^{1}\left ( \frac{x+1}{x^2+x+1}-\frac{x-1}{x^2-x+1}\right )dx$$=\frac{1}{4}\int\limits_{0}^{1}\left ( \frac{d(x^2+x+1)}{x^2+x+1}-\frac{d(x^2-x+1)}{x^2-x+1}\right )+\frac{1}{4}\int\limits_{0}^{1}\left ( \frac{1}{x^2+x+1}+\frac{1}{x^2-x+1}\right )dx$ $\left[ {\frac{1}{4}\ln\left\| {\frac{x^2+x+1}{x^2-x+1}}\right\|}+2\sqrt 3\arctan\frac{x-1}{\sqrt 3} +2\sqrt 3\arctan\frac{x-1}{\sqrt 3}\right]_0^1$

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