1. ## integration

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?

2. ## Re: integration

...what?

$\int \frac{1}{x} \, dx = \ln |x| + C$

3. ## Re: integration

no. try it through by parts method. then I is cancel out from both side.

4. ## Re: integration

Originally Posted by Swarnav
no. try it through by parts method. then I is cancel out from both side.
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 \begin{align*} \frac{d}{dx}\left(\ln{x}\right) = \frac{1}{x} \end{align*} for \displaystyle \begin{align*} x > 0 \end{align*} and \displaystyle \begin{align*} \frac{d}{dx}\left[\ln{(-x)}\right] = \frac{1}{x} \end{align*} for \displaystyle \begin{align*} x < 0 \end{align*}. Therefore

\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*}

5. ## Re: integration

Swarnav's right, letting $u = \frac{1}{x} \Rightarrow du = -\frac{1}{x^2}$ and $dv = dx \Rightarrow v = x$, you obtain

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

I might be wrong, but I'm 99.9% sure that the problem with this method is that you cannot assume that $dv = dx$ implies $v = x$. Maybe $v = -x$?

6. ## Re: integration

You're right, $dv=dx$ does not imply $v=x$ but $v=x+C$ 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:

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

In other words, you can't subtract $\int\frac{1}{x}\,dx$ from both sides, because the constants may be (and in this case are) different.

7. ## Re: integration

Thanks for helping me.

No problem