What is `int ((x^2 - 1)(x + 1))^(-2/3) dx`

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beckden eNotes educator| Certified Educator

`int ((x^2-1)(x+1))^(-2/3) dx`

This one is kind of tricky   `u=(x+1)/(x-1) du = -2/(x-1)^2 dx, -(x-1)^2/2 du = dx`

Since `(x^2-1) = (x+1)(x-1)` , 

`(x^2-1)(x+1) = (x+1)*(x+1)*(x-1) = (x+1)/(x-1)*(x+1)/(x-1)*(x-1)^3 = u^2(x-1)^3`

So we get 

`int ((x^2-1)(x+1))^(-2/3) dx = -int 1/((u^2(x-1)^3))^(2/3) * (x-1)^2/2 du =`

`= -int 1/((u^2)^(2/3)(x-1)^2)(x-1)^2/2 du = -1/2 int 1/u^(4/3) du =`

`= -1/2 int u^(-4/3) du = (-1/2)(-3) u^(-1/3) = 3/2 ((x+1)/(x-1))^(-1/3) + C`

So final answer is

` int ((x^2-1)(x+1))^(-2/3) dx = 3/2((x-1)/(x+1))^(1/3) + C`


lfryerda eNotes educator| Certified Educator

To find the integral `int(x^2-1)(x+1)^{-2/3}dx`, we need to use the substitution `u=x+1`.  This means that `du=dx`, so the integral becomes





`=3/5 u^{7/3}-3/2u^{4/3}+C` 

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