Improper Integrals questions - converge or diverge?

[zeion]

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

[A23]

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.