I'm sorry, I'm not sure what your question is. Could you clarify?

[twoflower]

Hi,

I'm trying to find this integral:

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

Because $1-x^2$ has two different real solutions, I can write

$$\sqrt{ax^2 + bx + c} = \sqrt{-a}(x_2 - x)\sqrt{\frac{x-x_1}{x_2 - x}}$$
so
$$\sqrt{1-x^2} = (-1 - x)\sqrt{\frac{x - 1}{-1 - x}} = (-1 - x)\sqrt{\frac{1-x}{1+x}}$$

I used this substitution:
$$t = \sqrt{\frac{1-x}{1+x}}$$

It gives
$$x = \frac{1-t^2}{1 + t^2}$$

$$x + 1 = \frac{2}{1+t^2}$$

$$dx = \frac{-4t}{(1+t^2)^2}$$

So
$$\int \frac{x}{(x+1)\sqrt{1-x^2}}\ dx = \int \frac{ \frac{1-t^2}{1+t^2}}{\left(\frac{2}{1+t^2}\right)\left(\frac{-2}{1+t^2}\right)t}\ \ \frac{-4t}{(1+t^2)^2}\ dt$$
$$= \int \frac{1-t^2}{1+t^2}\ dt = \int \frac{1}{1+t^2}\ dt - \int \frac{t^2}{1+t^2}\ dt = \arctan t -\ \int \frac{t^2}{1+t^2}\ dt$$

Damn I know I should be able to solve this integral, but I don't know how, maybe it's too late for me...

Btw the correct result should be:

$$\int \frac{x}{(x+1)\sqrt{1-x^2}}\ dx = \sqrt{\frac{1-x}{1+x}} + 2\arctan \sqrt{\frac{1+x}{1-x}} + C$$

Thank you.

 

[GCT]

Why don't you try a simple trig substitution and go from there? You can first get rid of the square root and try to simplify.

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

$$\int \frac{sin@cos@~d@}{(sin@cos@+cos@)}$$

Next try multiplying by $$\frac{(cos@-sin@cos@)}{cos@-sin@cos@}$$

 

[twoflower]

Thank you GCT, it's interesting idea, but it seems it leads to $t = \tan \frac{x}{2}$ substitution, which is little tricky in my experience...could you look please at my original solution and tell me what's wrong there?

Thanks!

 

[dextercioby]

A quick glance

$$x+1=\frac{2}{1+t^{2}}$$

,assuming "x" was right in the first place.

Daniel.

 

[twoflower]

A quick glance

$$x+1=\frac{2}{1+t^{2}}$$

,assuming "x" was right in the first place.

Daniel.

Thank you Daniel! How stupid I am..So I'm going to go solve it again, hopefully it will be ok now

 

[twoflower]

I edited my initial post now, could you now help me how to solve that?

 

[dextercioby]

$$\int \frac{t^{2}}{1+t^{2}} \ dt=\int \frac{1+t^{2}-1}{1+t^{2}} \ dt$$

Daniel.

 

[twoflower]

Thank you I noticed it already :) Now the result seems ok, except for sign and similar technicalities..

 

[dextercioby]

You can check the sign out.It looks okay to me.


Daniel.

 

[twoflower]

Well, the right result should be
$$\int \frac{x}{(x+1)\sqrt{1-x^2}}\ dx = \sqrt{\frac{1-x}{1+x}} + 2\arctan \sqrt{\frac{1+x}{1-x}} + C$$

but I have
$$2 \arctan \sqrt{\frac{1-x}{1+x}} - \sqrt{\frac{1-x}{1+x}}$$

 

[dextercioby]

I don't have time now,i'll take a deeper look later.

Daniel.

 

[Yegor]

$$\int \frac{x}{(x+1)\sqrt{1-x^2}}\ dx=\int \frac{x+1-1}{(x+1)\sqrt{1-x^2}}\ dx=\int \frac{1}{\sqrt{1-x^2}}\ dx-\int \frac{1}{(x+1)\sqrt{1-x^2}}\ dx$$
First one is obvious. For second $$t=\frac{1}{x+1}$$ will lead to simple solution.
The answer then is different from what you've given, but Mathematica says that it's absolutely correct.

 

[GCT]

Twoflower, in case you're interested

$$\int \frac{sin@cos@~d@}{(sin@cos@+cos@)} * \frac{(cos@-sin@cos@)}{cos@-sin@cos@}$$

$$\int \frac{sin@cos^{2}@-sin^{2}cos^{2}@~d@}{cos^{2}@-sin^{2}@cos^{2}@}$$

$$\int \frac{sin@cos^{2}@}{cos^{2}@-sin^{2}@cos^{2}@}~-~ \int \frac{sin^{2}@cos^{2}@~d@}{cos^{2}@-sin^{2}@cos^{2}@}$$

$$\int \frac{sin@~d@}{cos^{2}@}~-~ \int \frac{sin^{2}@~d@}{cos^{2}@}$$

The left term can be resolved using substitution. The right term simplifies to
$$\int \frac{1}{cos^{2}@}~-~\int d@$$

anyone see any mistakes? Please point them out.

 

[Yegor]

$$\int \frac{sin@~d@}{cos^{2}@} = -\int \frac{d(\cos@)}{cos^{2}@}$$

It's all right with your solution GCT. At least it's not so artificiall as which Twoflower made.