#I have done part a) where g(x) is just 1/f(x) but I don't understand how that helps with part b)?
quick nebula
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swift snow
Multiply the top and bottom of the integral by g(x). That will get rid of the denominator.
quick nebula
but won't the numerator just be (5-x)/(2-sqrt[x-1]) and it just goes in circles?
swift snow
but won't the numerator just be (5-x)/(2-sqrt[x-1]) and it just goes in circles?
No. We have:
(5 - x)/f(x) = (5 - x)g(x)/(f(x)g(x)) = (5 - x)g(x)
And if you calculated g(x) correctly, you'll see why it becomes easier.
quick nebula
wait but integral is (5-x)/f(x) so multiply both top/bottom by g(x) makes it (5-x)g(x) / f(x)g(x)=(5-x)/f(x) so nothing changes? Could you elaborate?
swift snow
wait but integral is (5-x)/f(x) so multiply both top/bottom by g(x) makes it (5-...
Um, no. Read what I just wrote.
Alright, first of all, what did you get for g(x)?
quick nebula
1/(2-sqrt[x-1])
reciprocal of f(x)
so the original intergal is (5-x)/f(x)=(5-x)g(x)
quick nebula
you mean like "rationalie the the denominatior"?
swift snow
Yes.
quick nebula
omw
(2+sqrt[x-1]) / (5-x)?
so just integral of (2+sqtrt[x-1]) after cancellations?
swift snow
Yeah.