1. ## Definite Integral 1/(x*(1-x)^0.5)

Hi I am trying to evaluate this integral using software and manually between 0 and 0.1 (or any upper limit) but I get Inf in the software solution and I haven't been able to solve it manually any help?

2. int 1/(x*(1-x)^0.5) - Wolfram|Alpha

I took a look at it and how do I evaluate my limits

int 1/(x*(1-x)^0.5) - Wolfram|Alpha

I took a look at it and how do I evaluate my limits
Use the substitution. From wolfram:

$u = 1-x$

Use this to find out limits in terms of u

$u = 1-0 = 1$

$u = 1-0.1 = 0.9$

So your limits are 1 and 0.9

4. Hi,

The problem I am having with that solution is that when I evaluate the limits I get imaginary numbers. And the goal of evaluating my integral is to find the area under the curve to obtain a critical factor that I am looking for. I need a solution that gives me real numbers.

Thanks again for all the help.

Hi,

The problem I am having with that solution is that when I evaluate the limits I get imaginary numbers. And the goal of evaluating my integral is to find the area under the curve to obtain a critical factor that I am looking for. I need a solution that gives me real numbers.

Thanks again for all the help.
from the wolfram graph of the integral plots for varying bounds, the integral is real for all $[0+\epsilon,1)$ where $\epsilon>0$

6. I appreciate all the help again,

Ok, So the reason why I am in need to solve this type of integral is because I need to integrate the product of 2 functions where the singularity point is very important in the physical problem I have. So since its the product of 2 functions (the one function is the one having singuarities and the integral being improper) then maybe I can use the mean value theorem to solve this problem; so, my next question is if the following is true?
$
\int_0^{a}m(r,a)s(r)\;dr =
$

$
\int_e^{a-e}m(r,a)s(r)\;dr +s(r)r=a\int_{a-e}^{a}m(r,a)\;dr +
$

$
m(r,a)r=e\int_{0}^{e}s(r)\;dr
$

$
e<$

My question is if I can use the mean value theorem on an integral of a product of two functions interchengeably even if my integral I divide it into sums of integrals and evaluate at the upper limit whatever function works for me and then multiply by the integral of the second function which id defined on the range integrated.

Thanks,

7. Sorry I made a mistake. For the last term $m(r,a)$ should be evaluated at $r=0$ instead of $r=e$