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

Nice solution!Twoflower, you should understand that the integral you was trying to solve doesn't exist.
  • #1
twoflower
368
0
Hi,

I'm trying to find this integral:

[tex]
\int \frac{x}{(x+1)\sqrt{1-x^2}}\ dx
[/tex]

Because [itex]1-x^2[/itex] has two different real solutions, I can write

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

I used this substitution:
[tex]
t = \sqrt{\frac{1-x}{1+x}}
[/tex]

It gives
[tex]
x = \frac{1-t^2}{1 + t^2}
[/tex]

[tex]
x + 1 = \frac{2}{1+t^2}
[/tex]

[tex]
dx = \frac{-4t}{(1+t^2)^2}
[/tex]

So
[tex]
\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
[/tex]
[tex]
= \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
[/tex]

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

Btw the correct result should be:

[tex]
\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
[/tex]

Thank you.
 
Last edited:
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  • #2
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.

[tex]
\int \frac{x}{(x+1)\sqrt{1-x^2}}\ dx
[/tex]

[tex] \int \frac{sin@cos@~d@}{(sin@cos@+cos@)} [/tex]

Next try multiplying by [tex] \frac{(cos@-sin@cos@)}{cos@-sin@cos@} [/tex]
 
  • #3
Thank you GCT, it's interesting idea, but it seems it leads to [itex]t = \tan \frac{x}{2}[/itex] substitution, which is little tricky in my experience...could you look please at my original solution and tell me what's wrong there?

Thanks!
 
  • #4
A quick glance

[tex] x+1=\frac{2}{1+t^{2}} [/tex]

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

Daniel.
 
  • #5
dextercioby said:
A quick glance

[tex] x+1=\frac{2}{1+t^{2}} [/tex]

,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 :smile:
 
  • #6
I edited my initial post now, could you now help me how to solve that?
 
  • #7
[tex] \int \frac{t^{2}}{1+t^{2}} \ dt=\int \frac{1+t^{2}-1}{1+t^{2}} \ dt [/tex]

Daniel.
 
  • #8
Thank you I noticed it already :) Now the result seems ok, except for sign and similar technicalities..
 
  • #9
You can check the sign out.It looks okay to me.


Daniel.
 
  • #10
Well, the right result should be
[tex]
\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
[/tex]

but I have
[tex]
2 \arctan \sqrt{\frac{1-x}{1+x}} - \sqrt{\frac{1-x}{1+x}}
[/tex]
 
  • #11
I don't have time now,i'll take a deeper look later.

Daniel.
 
  • #12
[tex]\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[/tex]
First one is obvious. For second [tex]t=\frac{1}{x+1}[/tex] will lead to simple solution.
The answer then is different from what you've given, but Mathematica says that it's absolutely correct.
 
  • #13
Twoflower, in case you're interested

[tex] \int \frac{sin@cos@~d@}{(sin@cos@+cos@)} * \frac{(cos@-sin@cos@)}{cos@-sin@cos@} [/tex]

[tex] \int \frac{sin@cos^{2}@-sin^{2}cos^{2}@~d@}{cos^{2}@-sin^{2}@cos^{2}@} [/tex]

[tex] \int \frac{sin@cos^{2}@}{cos^{2}@-sin^{2}@cos^{2}@}~-~ \int \frac{sin^{2}@cos^{2}@~d@}{cos^{2}@-sin^{2}@cos^{2}@}[/tex]

[tex] \int \frac{sin@~d@}{cos^{2}@}~-~ \int \frac{sin^{2}@~d@}{cos^{2}@} [/tex]

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

anyone see any mistakes? Please point them out.
 
  • #14
[tex]\int \frac{sin@~d@}{cos^{2}@} = -\int \frac{d(\cos@)}{cos^{2}@}[/tex]

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

1. What is an integral?

An integral is a mathematical concept that represents the area under a curve on a graph. It is an essential tool in calculus and is used to solve problems related to finding the total quantity or value of something.

2. What is a square root?

A square root is the number that, when multiplied by itself, gives the original number. For example, the square root of 9 is 3, because 3 multiplied by itself is 9.

3. How do you solve an integral with a square root?

To solve an integral with a square root, you first need to identify the function inside the square root. Then, you can use substitution or integration by parts to simplify the integral and solve for the area under the curve.

4. Can an integral with a square root have multiple solutions?

Yes, an integral with a square root can have multiple solutions. This is because there are various methods and techniques that can be used to solve integrals, and different methods may result in different solutions.

5. Why is it important to understand integrals with square roots?

Understanding integrals with square roots is important because they can be used to solve real-world problems, such as calculating the volume of irregular shapes or the work done by a variable force. It is also a crucial concept in calculus and is often used in higher-level math and science courses.

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