zeion Posted March 17, 2010 Report Share Posted March 17, 2010 Hello. I have some questions on this assignment, I'm wondering if I could get some help:Determine whether the integral converges and, if so, evaluate the integral. 1) [math] \int_{e}^{\infty} \frac{dx}{xlnx} [/math] I integrate and get [math] \lim_{b \to \infty} \int_{e}^{b} \frac{dx}{xlnx} = \lim_{ b \to \infty} [ \frac{ln(xlnx)}{lnx}]_{e}^{b} [/math] ? 2) [math] \int_{1}^{4} \frac{dx}{x^2 - 4} [/math] It has discontinuity at x = 2 and x = -2 so I evaluate [math] \int_{1}^{2} \frac{dx}{x^2 - 4} \int_{2}^{4} \frac{dx}{x^2 - 4} [/math] I integrate and get [math] \frac{-1}{4} \lim_{c \to \infty} \int_{1}^{c} \frac{1}{x+2} - \frac{1}{x-2} dx [/math] I sub in and get ln0? 3) [math]\int_{e}^{\infty} \frac{dx}{(lnx)^2}[/math] Does this integrate into [math] \lim_{b \to \infty} [\frac {-x}{lnx}]_{e}^{b} [/math] ? 4) [math] \int_{2}^{\infty} \frac {dx}{x^2 + sinx} [/math] I don't know how to integrate this Quote Link to comment Share on other sites More sharing options...

A23 Posted May 2, 2010 Report Share Posted May 2, 2010 For (1) I got : [math]\int \frac{1}{x\ln(x)}dx\underbrace{=}_{y=\ln(x),dy=dx/x}\int\frac{dy}{y}=\ln(y)=\ln(\ln(x))[/math] However, I don't know if it gives the same as your expression. (2) doesn't it give sthg. proto. : [math]\ln\left(\frac{x-2}{x+2}\right)[/math] ? (3) I derived : [math]\left(\frac{x}{\ln(x)}\right)'=\frac{\ln(x)-1}{\ln(x)^2}[/math] which looks near to the result, but it seems not to be the same. Quote Link to comment Share on other sites More sharing options...

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