Recent content by sinkdeep

I see. Sorry for making a big mess here.

\int{frac{1}{x\ln{x}}}\d x=\int{frac{1}{\ln{x}}}\d \ln{x}

[tex]\int{frac{1}{x\ln{x}}}\d x=\int{frac{1}{\ln{x}}}\d \ln{x}[\tex]

\int{frac{1}{x\ln{x}}}\d x=\int{frac{1}{\ln{x}}}\d \ln{x}

1/cos(x)dx =cos(x)/(1-sin^2(x))dx =1/(1-sin^2(x))d(sin(x))

If f(x) is continuous, I'd say there's no such func. Here's the reason: 1.The simplest case, that f(x) have no zero points in (0,1), so the curve is like bridge. For instance, let f(x)=sin(4*pi*x), and g(x)=f(x)-f(x+2/5), then g(0)<0, g(3/5)>0, and since g(x) is continuous, there should be...