1. ## Integration Problem

I would be very grateful if someone could help me with finding the integral for this little fellow...

1/ (x*((1-x)^0.5))

I have tried the few integration methods I know of (I've just finished A-Level), such as substitution and by parts, but they didn't appear to work. In addition, I put this into Wolfram's online integrator, and it came up with answer in a form I didn't understand, something to do with hypergeometric functions. I tried reading the article on them, but I couldn't see how to apply it to this problem. Are hypergeometric functions necessary for tackling this problem? They certainly look very confusing...

Jessica.

2. Originally Posted by j_clough
I would be very grateful if someone could help me with finding the integral for this little fellow...

1/ (x*((1-x)^0.5))

I have tried the few integration methods I know of (I've just finished A-Level), such as substitution and by parts, but they didn't appear to work. In addition, I put this into Wolfram's online integrator, and it came up with answer in a form I didn't understand, something to do with hypergeometric functions. I tried reading the article on them, but I couldn't see how to apply it to this problem. Are hypergeometric functions necessary for tackling this problem? They certainly look very confusing...

Jessica.
$\int \frac{1}{x \sqrt{1 - x}} \, dx = \ln |\sqrt{1 - x} - 1| - \ln |\sqrt{1 - x} + 1| + C$

I might have time later to give some details if no-one else has.

3. It's a standard form.

$(\text{arcsech(x)})' = \frac{-1}{x\sqrt{1-x^2}}$

Notice that in your integral, it is actually:
$\frac{1}{x\sqrt{1-(\sqrt{x})^2}}$

4. Originally Posted by j_clough
I would be very grateful if someone could help me with finding the integral for this little fellow...

1/ (x*((1-x)^0.5))

I have tried the few integration methods I know of (I've just finished A-Level), such as substitution and by parts, but they didn't appear to work. In addition, I put this into Wolfram's online integrator, and it came up with answer in a form I didn't understand, something to do with hypergeometric functions. I tried reading the article on them, but I couldn't see how to apply it to this problem. Are hypergeometric functions necessary for tackling this problem? They certainly look very confusing...

Jessica.
It's a standard solution but if you've got to show steps, you can do this. I've done this question and it's got a few steps so I'll explain instead of showing all steps. You can use substitution but not just one, it requires multiple substitution.

$\displaystyle\int \frac{1}{(x)(1-x)^{0.5}} \ \mathrm{d}x$

Substituting $u= \frac{1}{(1-x)^{0.5}}$ and doing the sorting out will lead to...

$2 \displaystyle\int \frac{1}{\left(1-\frac{1}{u^2}\right)u^2} \ \mathrm{d}u$

Here, substituting $v = \frac{1}{u^2}$ and doing the sorting out will lead to...

$-2 \displaystyle\int \frac{1}{1-v^2}\ \mathrm{d}v$

Now, just split into partial fraction and integrate. (You may require one more substitution when integrating a fraction).

Remember to substitute all variables ( $u,v ...$)to get the final answer in terms of $x$. The answer will be $\ln ((1 - x)^{0.5}-1) - \ln ((1 - x)^{0.5}+1) + c$

5. Originally Posted by Air
It's a standard solution but if you've got to show steps, you can do this. I've done this question and it's got a few steps so I'll explain instead of showing all steps. You can use substitution but not just one, it requires multiple substitution.

$\displaystyle\int \frac{1}{(x)(1-x)^{0.5}} \ \mathrm{d}x$

Substituting $u= \frac{1}{(1-x)^{0.5}}$ and doing the sorting out will lead to...

$2 \displaystyle\int \frac{1}{\left(1-\frac{1}{u^2}\right)u^2} \ \mathrm{d}u$

Here, substituting $v = \frac{1}{u^2}$ and doing the sorting out will lead to...

$-2 \displaystyle\int \frac{1}{1-v^2}\ \mathrm{d}v$

Now, just split into partial fraction and integrate. (You may require one more substitution when integrating a fraction).

Remember to substitute all variables ( $u,v ...$)to get the final answer in terms of $x$. The answer will be $\ln ((1 - x)^{0.5}-1) - \ln ((1 - x)^{0.5}+1) + c$
The substitution $1 - x = u^2$ is probably a little easier.

The reciprocal substitution $x = \frac{1}{u}$ can also be used and leads to the standard result quoted by chops.

It's not difficult to show that the two answers are equivalent.

Originally Posted by Chop Suey
It's a standard form.

$(\text{arcsech(x)})' = \frac{-1}{x\sqrt{1-x^2}}$

Notice that in your integral, it is actually:

$\frac{1}{x\sqrt{1-(\sqrt{x})^2}}$

6. Ahh... je comprends! I'll remember this double substitution technique.

Thank you very much for all your replies.

Jessica.