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Mathematics 22 Online
OpenStudy (anonymous):

Integration Help Guys!!!

OpenStudy (anonymous):

OpenStudy (anonymous):

@amistre64

OpenStudy (anonymous):

@UCLAfan1991

OpenStudy (anonymous):

@thomaster

OpenStudy (anonymous):

@ankit042

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!!

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

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

OpenStudy (anonymous):

No Problem

OpenStudy (anonymous):

@Jamierox4ev3r ......NP on dat....who cares!!!that was just chat talk!!Tnx again!!

OpenStudy (anonymous):

@Jamierox4ev3r

HanAkoSolo (jamierox4ev3r):

lolz okay :P

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