Tricky? Definate Integrals

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March 3rd 2008, 04:39 PM
polymerase

Tricky? Definate Integrals

$\int^5_1 \frac{\ln x}{x^2+2x+5}\:dx$

Thanks
March 3rd 2008, 09:06 PM
Aryth

Some things are best left to the calculator:

$\int_1^5 \frac{ln(x)}{x^2 + 2x + 5}dx = 0.187 \ \text{units}^2$
March 3rd 2008, 09:13 PM
mr fantastic

Quote:

Originally Posted by polymerase

$\int^5_1 \frac{\ln x}{x^2+2x+5}\:dx$

Thanks

Do you have reason to think that an exact solution exists?
March 4th 2008, 04:13 AM
polymerase

Quote:

Originally Posted by mr fantastic

Do you have reason to think that an exact solution exists?

This was a past test question....we are not allowed to use calculators!....so there is definalty a way to find the exact answer. I just don't know how.

If u don't believe me, the guy said it was "0.187".....plug $\frac{\ln5}{4}\left(\arctan3-\frac{\pi}{4}\right)$ and tell me what you get :D
March 4th 2008, 04:38 AM
Demen

Using
x = z-1

and
z = 2 tan Θ

I got
∫ ln (2tan Θ - 1) dΘ

but I'm not exactly sure what the next step may be.
March 4th 2008, 01:38 PM
mr fantastic

Quote:

Originally Posted by polymerase

This was a past test question....we are not allowed to use calculators!....so there is definalty a way to find the exact answer. I just don't know how.

If u don't believe me, the guy said it was "0.187".....plug $\frac{\ln5}{4}\left(\arctan3-\frac{\pi}{4}\right)$ and tell me what you get :D

No elementary primitive exists, so it has to be handled as a definite integral from the word go. That implies special teachniques for example contour integration.

So what we need to know is:

What sort of mathematics is this question contexted in? The answer to this question will suggest the technique to use.
March 4th 2008, 03:28 PM
Krizalid

Quote:

Originally Posted by polymerase

$\int^5_1 \frac{\ln x}{x^2+2x+5}\:dx$

Thanks

Let $\lambda = \int_1^5 {\frac{{\ln x}}<br /> {{x^2 + 2x + 5}}\,dx}.$

After substitution $x=\frac5u,$ we have $\lambda = \int_1^5 {\frac{{\ln 5 - \ln x}}<br /> {{x^2 + 2x + 5}}\,dx}.$

Hence $\lambda = \frac{1}<br /> {2}\ln 5\int_1^5 {\frac{1}<br /> {{x^2 + 2x + 5}}\,dx}.$ The rest follows.