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

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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

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

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

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