Calculate the generalized integral

**sf1903** · December 15th 2009, 11:13 AM

Am I correct? and then?

**tonio** · December 15th 2009, 11:33 AM

Originally Posted by sf1903

Calculate the generalized integral

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Am I correct? and then?

Sorry to say but you're not even close to correct, and if you wrote the above then I think you need to go back to your general basics on integration:

$\int\frac{x^2-1}{x^4}dx=\int\left(\frac{1}{x^2}-\frac{1}{x^4}\right)dx=-\frac{1}{x}+\frac{1}{3x^3}$ .

Now evaluate the above in the integral's limits and make the upper limit tend to $\infty$ ...the answer must be $\frac{2}{3}$

Tonio

**Krizalid** · December 15th 2009, 12:48 PM

we can also turn this improper integral into a proper one, so let's put $x=\frac1t$ and the integral becomes $\int_{0}^{1}{\left( 1-t^{2} \right)\,dt}=1-\frac{1}{3}=\frac{2}{3}.$

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