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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: