1. ## Tricky? Definate Integrals

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

Thanks

2. Some things are best left to the calculator:

$\displaystyle \int_1^5 \frac{ln(x)}{x^2 + 2x + 5}dx = 0.187 \ \text{units}^2$

3. Originally Posted by polymerase
$\displaystyle \int^5_1 \frac{\ln x}{x^2+2x+5}\:dx$

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

4. 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 $\displaystyle \frac{\ln5}{4}\left(\arctan3-\frac{\pi}{4}\right)$ and tell me what you get

5. 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.

6. 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 $\displaystyle \frac{\ln5}{4}\left(\arctan3-\frac{\pi}{4}\right)$ 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.

7. Originally Posted by polymerase

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

Thanks
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.