archcommus
Diamond Member
- Sep 14, 2003
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Originally posted by: Engineer
Calculus II.
Calculus III and D.E. were both easier.
There are lots of ways to solve PDEs... They're just not happy.Originally posted by: KillerCharlie
I thought Calc 2 was harder than those as well. I'm specializing in aerodynamics so vector calculus is pretty straightforward and easy to apply.
I was going to take a math course in partial differential equations... where all they do is tell you they can't be solved. Instead I took a course in computational fluid dynamics and went about solving them... numerically of course.
Originally posted by: archcommus
All I know is I keep screwing up my integration by parts problems!!
What is it?Originally posted by: Soccer55
Originally posted by: archcommus
All I know is I keep screwing up my integration by parts problems!!
If you're having a problem figuring out what u and dv should be, use LIPET.
-Tom
Originally posted by: Astaroth33
It's been a long time, but I'd say the calculus one where you do integrals and derivatives of infinite series.
Originally posted by: archcommus
Okay I cannot integrate ln (2x+1) dx for the life of me. Dammit what am I doing wrong.
Originally posted by: Random Variable
Originally posted by: archcommus
Okay I cannot integrate ln (2x+1) dx for the life of me. Dammit what am I doing wrong.
integrate by parts
Originally posted by: Soccer55
Originally posted by: Random Variable
Originally posted by: archcommus
Okay I cannot integrate ln (2x+1) dx for the life of me. Dammit what am I doing wrong.
integrate by parts
No, you don't need integration by parts. Try a regular u substitution with u=2x+1.
-Tom
Originally posted by: Random Variable
Originally posted by: Soccer55
Originally posted by: Random Variable
Originally posted by: archcommus
Okay I cannot integrate ln (2x+1) dx for the life of me. Dammit what am I doing wrong.
integrate by parts
No, you don't need integration by parts. Try a regular u substitution with u=2x+1.
-Tom
That will work only if you know that the integral of ln(x)= x*ln(x)-x