Let (I=Integration of 1/x dx). break 1/x as ((1/x)*1),then integrate it. then we get 1=0. Where is wrong?

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- Jul 7th 2012, 10:02 AMSwarnavintegration
Let (I=Integration of 1/x dx). break 1/x as ((1/x)*1),then integrate it. then we get 1=0. Where is wrong?

- Jul 7th 2012, 10:19 AMrichard1234Re: integration
...what?

$\displaystyle \int \frac{1}{x} \, dx = \ln |x| + C$ - Jul 7th 2012, 10:43 PMSwarnavRe: integration
no. try it through by parts method. then I is cancel out from both side.

- Jul 7th 2012, 11:20 PMProve ItRe: integration
That's giving you a big hint that Integration by Parts is an inappropriate method to use in this case. It's known that $\displaystyle \displaystyle \begin{align*} \frac{d}{dx}\left(\ln{x}\right) = \frac{1}{x} \end{align*}$ for $\displaystyle \displaystyle \begin{align*} x > 0 \end{align*}$ and $\displaystyle \displaystyle \begin{align*} \frac{d}{dx}\left[\ln{(-x)}\right] = \frac{1}{x} \end{align*}$ for $\displaystyle \displaystyle \begin{align*} x < 0 \end{align*}$. Therefore

$\displaystyle \displaystyle \begin{align*} \int{\frac{1}{x}\,dx} &= \begin{cases} \ln{x} + C \textrm{ if }x > 0 \\ \ln{(-x)}+ C\textrm{ if } x < 0 \end{cases} \\ &= \ln{|x|} + C \end{align*}$ - Jul 8th 2012, 12:03 AMrichard1234Re: integration
Swarnav's right, letting $\displaystyle u = \frac{1}{x} \Rightarrow du = -\frac{1}{x^2}$ and $\displaystyle dv = dx \Rightarrow v = x$, you obtain

$\displaystyle \int \frac{1}{x} \, dx = 1 - \int -\frac{1}{x} \Rightarrow 1 = 0$

I might be wrong, but I'm 99.9% sure that the problem with this method is that you cannot assume that $\displaystyle dv = dx$ implies $\displaystyle v = x$. Maybe $\displaystyle v = -x$? - Jul 8th 2012, 12:49 AMthesmurfmasterRe: integration
You're right, $\displaystyle dv=dx$ does not imply $\displaystyle v=x$ but $\displaystyle v=x+C$ for a constant C. When you integrate by parts you usually don't write this constant out in the intermediate steps because you still have indefinite integrals on both sides. To be strict, however, you should write:

$\displaystyle \int\frac{1}{x}\, dx=1+C-\int-\frac{1}{x}\, dx$

In other words, you can't subtract $\displaystyle \int\frac{1}{x}\,dx$ from both sides, because the constants may be (and in this case are) different. - Jul 8th 2012, 01:43 AMSwarnavRe: integration
Thanks for helping me.

- Jul 8th 2012, 05:52 PMrichard1234Re: integration
No problem :)