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OpenStudy (anonymous):
OpenStudy (anonymous):
@amistre64
OpenStudy (anonymous):
@UCLAfan1991
OpenStudy (anonymous):
@thomaster
OpenStudy (anonymous):
@ankit042
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OpenStudy (anonymous):
I am trying to remember how to do this
OpenStudy (anonymous):
@Preetha
OpenStudy (anonymous):
|dw:1375805220245:dw|
OpenStudy (psymon):
Let me give it a shot in a sec, lol.
OpenStudy (anonymous):
Nope...M posting the answer!!
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OpenStudy (psymon):
Alright O.o
OpenStudy (anonymous):
Sorry Cambrige I was throwing out my guess out there
OpenStudy (anonymous):
NP UCLA
OpenStudy (anonymous):
The answer is this....
OpenStudy (anonymous):
I told you its been awhile since I seen these problems
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OpenStudy (psymon):
Yeah, what you got looks right.
u = bx^n + c. so du = nbx^(n-1)dx\[\frac{ ax ^{n-1} }{ u }*\frac{ du }{ bnx ^{n-1} }\]
x^(n-1) cancels out and I can factor a/bn out of the integral
\[\frac{ a }{ bn }\int\limits_{}^{}\frac{ 1 }{ u }\]
\[\frac{ a }{ bn }*\ln (u) + k = \frac{ a }{ bn }\ln (bx ^{n}+ c) + k\]
OpenStudy (anonymous):
Thanks @Psymon Appreciate it!!
OpenStudy (psymon):
Yep, np ^_^
OpenStudy (anonymous):
I tried
OpenStudy (anonymous):
Ya tnx for tryin UCLA
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OpenStudy (anonymous):
No Problem
OpenStudy (anonymous):
@Jamierox4ev3r ......NP on dat....who cares!!!that was just chat talk!!Tnx again!!