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

Thanks

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- Mar 3rd 2008, 04:39 PMpolymeraseTricky? Definate Integrals
$\displaystyle \int^5_1 \frac{\ln x}{x^2+2x+5}\:dx$

Thanks - Mar 3rd 2008, 09:06 PMAryth
Some things are best left to the calculator:

$\displaystyle \int_1^5 \frac{ln(x)}{x^2 + 2x + 5}dx = 0.187 \ \text{units}^2$ - Mar 3rd 2008, 09:13 PMmr fantastic
- Mar 4th 2008, 04:13 AMpolymerase
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 $\displaystyle \frac{\ln5}{4}\left(\arctan3-\frac{\pi}{4}\right)$ and tell me what you get :D - Mar 4th 2008, 04:38 AMDemen
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. - Mar 4th 2008, 01:38 PMmr fantastic
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. - Mar 4th 2008, 03:28 PMKrizalid
Let $\displaystyle \lambda = \int_1^5 {\frac{{\ln x}}

{{x^2 + 2x + 5}}\,dx}.$

After substitution $\displaystyle x=\frac5u,$ we have $\displaystyle \lambda = \int_1^5 {\frac{{\ln 5 - \ln x}}

{{x^2 + 2x + 5}}\,dx}.$

Hence $\displaystyle \lambda = \frac{1}

{2}\ln 5\int_1^5 {\frac{1}

{{x^2 + 2x + 5}}\,dx}.$ The rest follows.