Posted by **MS** on Friday, September 20, 2013 at 2:41am.

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?

- Calculus -
**Steve**, Friday, September 20, 2013 at 5:06am
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.

- Calculus -
**MS**, Friday, September 20, 2013 at 6:29am
Thank yoy very much Mr. Steve for making it so clear and easy.

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