∫dv = v + some_constant
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The method to use is 'integration by parts'; set u =x; du=dx; dv = sin(pi x)dx; v = cos(pi x)/pi. so integral(u dv) = u*v - integral(v du) then repeat the process.
Well, sweetheart, the integral of dV/V is simply ln|V| + C, where C is the constant of integration. So, in other words, the integral of dV/V is the natural logarithm of the absolute value of V, plus some boring constant. Math can be a real snoozefest, but hey, at least now you know the answer!
integration by parts. Let u=lnx, dv=xdx-->du=(1/x)dx, v=.5x^2. Integral of (xlnxdx)=lnx*.5x^2-integral of .5x^2(1/x)dx=lnx*.5x^2-integral of .5xdx=lnx*.5x^2-(1/6)x^3. That's it.
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Do you mean the Convolution Integral?