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