Integration problem with u as the variable

[Lizzy]

Are you suppose to treat this problem as if it was another variable like x or t?

The integral of 1/ [u^2(1+ u^2)] du

Sorry about this, but I also integrated this problem:

The integral of (x^5)(ln x)^2 dx

I got this:

x^6 [3(ln x)^2 -(1)] + C

Is this right?

Thank you soooo much for the help!!!

 

[galactus]

Are you suppose to treat this problem as if it was another variable like x or t?

The integral of 1/ [u^2(1+ u^2)] du

Maybe use partial fractions:

\(\displaystyle \L\\\frac{1}{u^{2}+u^{4}}=\frac{1}{u^{2}}-\frac{1}{u^{2}+1}\)

 

[trojan]

Hello. My first time replying to a question.

Yes you should treat the variable "u" in reference to "du" just like you would treat the variable "x" when you see a "dx" during integration.

To answer your second question, I don't think your answer is correct. I think you need to integrate by parts and choose u = lnx and dv = x^5 dx. Then you should be able to solve easily.

 

[galactus]

Trojan is correct.

\(\displaystyle \L\\u=(ln(x))^{2}, \;\ dv=x^{5}dx, \;\ du=\frac{2ln(x)}{x}dx, \;\ v=\frac{x^{6}}{6}\)