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Calculus

posted by on .

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 - ,

    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 - ,

    Thank yoy very much Mr. Steve for making it so clear and easy.

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