integration

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July 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?
July 7th 2012, 10:19 AM
richard1234

Re: integration

...what?

$\int \frac{1}{x} \, dx = \ln |x| + C$
July 7th 2012, 10:43 PM
Swarnav

Re: integration

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

$\displaystyle \begin{align*} \int{\frac{1}{x}\,dx} &= \begin{cases} \ln{x} + C \textrm{ if }x > 0 \\ \ln{(-x)}+ C\textrm{ if } x < 0 \end{cases} \\ &= \ln{|x|} + C \end{align*}$
July 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 $v = -x$ ?
July 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.
July 8th 2012, 01:43 AM
Swarnav

Re: integration

Thanks for helping me.
July 8th 2012, 05:52 PM
richard1234

Re: integration

No problem :)