How do I integrate this? 1/(x^2-1)^.5
How do I integrate this? x/(x^2-1)^.5
For the first, make a u-substitution.
For the second, make a trigonometric substitution suggested by that difference of squares.
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).
So your integral becomes:
∫dx/√(x2-1) = ∫secødø = ln|secø+tanø|
Substituting back in
Think about that some more
By u substitution, let u=x^2-1 then du=2x
you have x you only need 2
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