Asked by mathstudent
Integrate: y/sqrt(2y+1) dx
Answers
Answered by
drwls
You must mean y/sqrt(2y+1) dy
Use the substitution 2y + 1 = u
dy = du/2
y = (u-1)/2
The integral becomes
(Integral of) [(1/2)(u-1)/sqrt u]du/2
= (Integral of) (1/4) [sqrt u - 1/sqrt u] du
You finish it.
Use the substitution 2y + 1 = u
dy = du/2
y = (u-1)/2
The integral becomes
(Integral of) [(1/2)(u-1)/sqrt u]du/2
= (Integral of) (1/4) [sqrt u - 1/sqrt u] du
You finish it.
Answered by
mathstudent
Thanks!
Actually, the problem was printed in my textbook like that with the equation using the variable y, but with dx rather than dy.
This seems to be a textbook error. I wasn't sure whether that was the case or whether I was doing something wrong.
Actually, the problem was printed in my textbook like that with the equation using the variable y, but with dx rather than dy.
This seems to be a textbook error. I wasn't sure whether that was the case or whether I was doing something wrong.
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