- #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...
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.
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...
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: