Integral of (3x - 1) / (3x)

[koreamaniac101]

To begin, I know how to correctly integrate this problem:

∫(3x - 1 ∕ 3x) dx = ∫(1 - 1/3x) dx
= ∫(1 - (1/3 * 1/x)) dx

= x - 1/3 * ln(x)



But in my initial attempt to solve the problem, I tried converting 1/3x to 3x^-1, thus changing the integral to be

= ∫(1 - 3x^-1) dx
= x - 3ln(x)
However, I was told that I cannot convert 1/3x to 3x^-1. This may be more of an algebraic question, but why can this conversion not occur?


-Soohan

 

[MarkFL]

What you want is:

\(\displaystyle \displaystyle \frac{1}{3x}=(3x)^{-1}\)

Note that:

\(\displaystyle \displaystyle 3x^{-1}=\frac{3}{x}\)

 

[koreamaniac101]

What you want is:

\(\displaystyle \displaystyle \frac{1}{3x}=(3x)^{-1}\)

Note that:

\(\displaystyle \displaystyle 3x^{-1}=\frac{3}{x}\)

Ah, I understand. So with (3x)^-1, how would I integrate the problem? I can see that u-substitution gives me the correct answer, but for curiosity's sake, is there another way to integrate 3x^-1?

 

[MarkFL]

Ah, I understand. So with (3x)^-1, how would I integrate the problem? I can see that u-substitution gives me the correct answer, but for curiosity's sake, is there another way to integrate 3x^-1?

I would likely write:

\(\displaystyle \displaystyle \int \frac{3x-1}{3x}\,dx=\int 1-\frac{1}{3}x^{-1}\,dx=x-\frac{1}{3}\ln|x|+C\)