Integrate (xdx)/(sqrt(x^2-1)) ?

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Answer 1 · 2018-04-10T21:04:02

#intx/sqrt(x^2-1)dx=sqrt(x^2-1)+C#

.

#intx/sqrt(x^2-1)dx#

Let #u=x^2-1#

#du=2xdx#

#xdx=(du)/2#

Let's substitute:

#int1/sqrtu*(du)/2=1/2intu^(-1/2)du=1/2(2)u^(1/2)=sqrtu+C#

Now, we can substitute back:

#intx/sqrt(x^2-1)dx=sqrt(x^2-1)+C#