How do you test the improper integral #int 2x(x^2-1)^(-1/3)dx# from #[0,1]# and evaluate if possible?

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Answer 1 · 2017-08-05T19:27:32

#int_0^1 2x(x^2-1)^(-1/3) dx = -3/2#

By substitution #int2x(x^2-1)^(-1/3) dx = 3/2(x^2-1)^(2/3) +C#

So,

#int_0^1 2x(x^2-1)^(-1/3) dx = lim_(brarr1^-)int_0^b 2x(x^2-1)^(-1/3) dx #

# = lim_(brarr1^-)[3/2(x^2-1)^(2/3)]_0^b#

# = lim_(brarr1^-)[3/2(b^2-1)^(2/3)]-[3/2(-1)^(2/3)]#

# = 0-3/2 = -3/2#