How to solve this integral #int 3/(x+1)^2 sqrt((x-1)/(x+1)) \ dx#?

#3/(x+1)^2##sqrt((x-1)/(x+1))#

1 Answer
Jun 12, 2018

Please see below.

Explanation:

Seeing #(x+1)^2# in the denominator of one factor and the other involving #(x-1)/(x+1)# it makes sense to consider the possibility of using a substitution. This is suggested by the fact that the derivative of #(x-1)/(x+1)# has denominator #(x+1)^2#.

Let #u = (x-1)/(x+1)#, then #du = (1(x+1)-(x-1)(1))/(x+1)^2 dx = 2/(x+1)^2 dx#

#int 3/(x+1)^2 sqrt((x-1)/(x+1)) \ dx = 3/2 int underbrace(((x-1)/(x+1))^(1/2))_(u^(1/2)) underbrace(2/(x+1)^2 dx)_(du)#

# = 3/2int u^(1/2) du = 3/2 2/3 u^(3/2) +C = u^(3/2) +C#

# = ((x-1)/(x+1))^(3/2) +C#

Check this answer by differentiating to make sure we get the original integrand.