# Integral with square root

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.

Last edited:

## Answers and Replies

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GCT
Homework Helper
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 $$\frac{(cos@-sin@cos@)}{cos@-sin@cos@}$$

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
Homework Helper
A quick glance

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

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

Daniel.

dextercioby said:
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 gonna go solve it again, hopefully it will be ok now

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

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

Daniel.

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

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

Daniel.

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

Daniel.

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

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

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