both are correct
log ax = log a + log x
so, the loga/a is included in the C when you do it. The book probably did it their way to retain the appearance of (ax) as a function of x.
int(1/ax) dx let u = ax and you have
1/a int 1/u du
integrate that to get 1/a logu + C
= 1/a log(ax) + C
The C's are different, that's all.
FWIW, I prefer your way.
I have done Int(1/ax)dx=1/a*Int(1/x)dx
=[(log x)/a] + C.
The working in the book is
Int(1/ax)dx=1/a*Int(a/ax)dx
=[(log ax)/a]+C.
Which one is correct and where is the error in my working, if so?
2 answers
Thank yoy very much Mr. Steve for making it so clear and easy.
2 answers