# integration

• Jul 7th 2012, 10:02 AM
Swarnav
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?
• Jul 7th 2012, 10:19 AM
richard1234
Re: integration
...what?

$\int \frac{1}{x} \, dx = \ln |x| + C$
• Jul 7th 2012, 10:43 PM
Swarnav
Re: integration
no. try it through by parts method. then I is cancel out from both side.
• Jul 7th 2012, 11:20 PM
Prove It
Re: integration
Quote:

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*}
• Jul 8th 2012, 12:03 AM
richard1234
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 $v = -x$?
• Jul 8th 2012, 12:49 AM
thesmurfmaster
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.
• Jul 8th 2012, 01:43 AM
Swarnav
Re: integration
Thanks for helping me.
• Jul 8th 2012, 05:52 PM
richard1234
Re: integration
No problem :)