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?
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
$\displaystyle \displaystyle \begin{align*} \int{\frac{1}{x}\,dx} &= \begin{cases} \ln{x} + C \textrm{ if }x > 0 \\ \ln{(-x)}+ C\textrm{ if } x < 0 \end{cases} \\ &= \ln{|x|} + C \end{align*}$
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 $\displaystyle v = -x$?
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.