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`int(xdx)/((x-1)(x-2))`は
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how Richard Feynman would integrate 1/(1+x^2)^2
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f(x)=e^x (0<=x<1), e^2-x (1<=x<=2) (0<=x<=2) g(x)=∫0toxlf(x)-f(t)ldt, g(4/3)=?
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有理関数 `(1-x^2)/(x(1-2x)` を積分します...
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Find `int(dx)/((x+1)(x+2)`...