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OpenStudy (anonymous):
Integrate. 1/sqrt(x^2+16)
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OpenStudy (ash2326):
We have
\[\int \frac{1}{\sqrt{x^2+16}} dx\]
Do you know integration by substitution?
OpenStudy (anonymous):
yeah, I think did something wrong because I got \[1/4\ln \left| \sqrt{x^2+16} =x\right|\]
OpenStudy (ash2326):
Could you share your work here?
OpenStudy (anonymous):
I think I know why it's raised to the 1/4 power right. I think I get it. because if everything it raised to that the numbers that are over 4 go away.
OpenStudy (anonymous):
you did u sub? did you learn trig subs?
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OpenStudy (anonymous):
Yes I did, I think I know my error.
OpenStudy (ash2326):
Could you try again? pr you want me to guide you?
@CatLove9
OpenStudy (ash2326):
*or
OpenStudy (anonymous):
I got it, I forgot about sec^2
OpenStudy (ash2326):
@CatLove9 so what did you get this time?
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OpenStudy (anonymous):
ln|sqrt(x^2+16)+x|
OpenStudy (ash2326):
there is a 1/4 missing in the integral. Would you check again?
OpenStudy (anonymous):
No there isn't I checked the answer in the book it's right because the dtheta is 4sec^2
OpenStudy (ash2326):
got it, sorry :(
OpenStudy (anonymous):
It's okay
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