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

Answer 1

Elijah Abram

#(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#

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Answer 2

Emilia Aguilar

The antiderivative of sqrt[(X – 1)/(X^5)] with respect to X is given by:

∫sqrt[(X – 1)/(X^5)] dX = ∫(X^(-5/2) - X^(-7/2)) dX = -2/3X^(-3/2) + 2/5X^(-5/2) + C, where C is the constant of integration.

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.