What is the antiderivative of #sqrt[(X – 1)/(X^5)]d#?

Question

8252 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-03-26T13:58:30

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

You are asking for:

#intsqrt((x-1)/x^5)dx#

We should simplify this by trying to remove squared terms. The first step is to split the #x^5# into #x^4(x)#.

#intsqrt((x-1)/(x^4(x)))dx=intsqrt(1/x^4((x-1)/x))dx#

Now, #sqrt(1/x^4)=1/x^2# so this can be brought out from under the square root.

#=int1/x^2sqrt((x-1)/x)dx#

Split up the fraction inside the square root.

#=int1/x^2sqrt(1-1/x)dx#

We can now use substitution--notice that we have some inner derivatives going on:

Let #u=1-1/x#. This also implies that #du=1/x^2dx#.

Substituting, we see now that

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

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