How do I integrate this? x/(x^2-1)^.5

[ComFlu945]

How do I integrate this? 1/(x^2-1)^.5

How do I integrate this? x/(x^2-1)^.5

And this
1/(x^2-1)^.5

 

[g_edgar]

For the first, make a u-substitution.

For the second, make a trigonometric substitution suggested by that difference of squares.

 

[ComFlu945]

Thanks!

First one worked like a charm.

For second one I substituted x for cosh(y). Since cosh(y)^2-1=sinh(y)^2, but bottom turns into sinh(y). And since x=cosh(y), dx/dy= sinh(y).

Back to original equation:
integral( 1/(x^2-1)^.5 dx) = integral ( sinh(y)/sinh(y)) dy = 1 + constant. However, the answer is supposed to be cosh^-1(x).

 

[Matthollyw00d]

Let x=secø
dx=secøtanødø
tanø=√(x²-1)

So your integral becomes:
∫dx/√(x²-1) = ∫secødø = ln|secø+tanø|

Substituting back in
ln|secø+tanø|=ln|x+√(x²-1)|

 

[g_edgar]

integral ( sinh(y)/sinh(y)) dy = 1 + constant

Think about that some more

 

[darkmagic]

By u substitution, let u=x^2-1 then du=2x
you have x you only need 2