How to Integrate x*ln(2x) and x^2/(x^2-4)?

[lastdayx52]

a. <br /> \int x*ln(2x) dx <br />
b. <br /> \int \frac{x^2}{x^2-4} dx <br />

My Attempt:
a. I have abs. no clue how to start it off. Do you start it off using integration by parts?
b. Same with this. I don't know how to start it off.

"Random" Questions:
1) What does "converge" and "diverge" mean? i.e. Another question asks "Determine if each integral converges or diverges". I can do the integrals, but I'm not sure what its asking. One results in the answer being pi/3, the other results in infinity.
2) When the question tells you to setup an integral for a volume of a solid generated when the region is rotated about the line x=-2 using cylindrical shells, you just do:
\pi *<br /> \int (f(x)+2)^2 dx <br />
Correct?

Thanks!

P.S. I'm not asking for answers, but rather how to start them off. Maybe 1 or 2 beginning steps in integrating these? =D

 

[gabbagabbahey]

Hi lastdayx52, welcome to PF

Let's start with a) and b)... For a) try integration by parts...For b), hint: x^2=(x^2-4)+4

Give it a shot and post your attempt if you require further assistance.

 

[lastdayx52]

Hi lastdayx52, welcome to PF

Let's start with a) and b)... For a) try integration by parts...For b), hint: x^2=(x^2-4)+4

Give it a shot and post your attempt if you require further assistance.

Got a:
\frac{x^2*(2*ln(2)-1)}{4}+\frac{x^2*ln(x)}{2} + C
Checked on calculator, and was correct. Thank you!

Now for b, that doesn't help me at all... >.>

 

[gabbagabbahey]

Now for b, that doesn't help me at all... >.>

Why not?

\int \frac{x^2}{x^2-4} dx=\int \frac{(x^2-4)+4}{x^2-4} dx=\int \frac{x^2-4}{x^2-4} dx +\int \frac{4}{x^2-4} dx

 

[lastdayx52]

Hi lastdayx52, welcome to PF

Let's start with a) and b)... For a) try integration by parts...For b), hint: x^2=(x^2-4)+4

Give it a shot and post your attempt if you require further assistance.

Why not?

\int \frac{x^2}{x^2-4} dx=\int \frac{(x^2-4)+4}{x^2-4} dx=\int \frac{x^2-4}{x^2-4} dx +\int \frac{4}{x^2-4} dx

LOL haha... See I did that, but stupidly thought the integral of 1 = 1, and when checked against the calculator, it was obviously not correct. haha... Thanks! =P