How do you evaluate the integral #int x/(sqrt(x-1)-sqrtx)#?

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Answer 1 · 2017-01-29T00:51:51

#-2/5(x-1)^(5/2)-2/3(x-1)^(3/2)-2/5x^(5/2)+C#

Multiply first by the conjugate of the denominator to simplify the integrand.

#I=int(x(sqrt(x-1)+sqrtx))/((sqrt(x-1)-sqrtx)(sqrt(x-1)+sqrtx))dx#

The denominator is now in the form #(a-b)(a+b)=a^2-b^2#, where #a=sqrt(x-1)# and #b=x#.

#I=int(x(sqrt(x-1)+sqrtx))/((x-1)-x)dx#

#I=-intx(sqrt(x-1)+sqrtx)dx#

Distributing, splitting up the integral and rewriting:

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

The second integral can be directly integrated using #intx^ndx=x^(n+1)/(n+1)+C#, where #n!=-1#.

#I=-intx(x-1)^(1/2)dx-x^(5/2)/(5/2)#

#I=-intx(x-1)^(1/2)dx-2/5x^(5/2)#

For the remaining integral, let #u=x-1#. This implies that #x=u+1# and #du=dx#.

#I=-int(u+1)u^(1/2)du-2/5x^(5/2)#

Distributing #(u+1)u^(1/2)# into separate integrals:

#I=-intu^(3/2)du-intu^(1/2)du-2/5x^(5/2)#

Using the rule from earlier:

#I=-u^(5/2)/(5/2)-u^(3/2)/(3/2)-2/5x^(5/2)+C#

From #u=x-1#:

#I=-2/5(x-1)^(5/2)-2/3(x-1)^(3/2)-2/5x^(5/2)+C#