How do you evaluate the definite integral #int (2x)/sqrt(x^2-1)# from 1 to sqrt5?

Question

14478 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-03-09T10:54:26

The answer is #=4#

We perform this integral by substitution

Let #u=x^2-1#

#du=2xdx#

Therefore,

#int(2xdx)/sqrt(x^2-1)#

#=int(du)/sqrtu#

#=2sqrtu#

So,

#int_1^sqrt5(2xdx)/sqrt(x^2-1)=[2sqrt(x^2-1)]_1^sqrt5#

#=2*2-0#

#=4#