indefinite integral of sqrt[ (x - 1) / x^5 ] dx

[k3232x]

I've been fooling around with this integral for about an hour and i am not getting anywhere. Mainly, I need help on where to start (ie. what substitution to make)

 

[stapel]

According to The Integrator, the integral is:

. . . . .\(\displaystyle \L \int\, \sqrt{\frac{x\, -\, 1}{x^5}\,}\, dx\, =\, \frac{2}{3}\, \left(\,\frac{x\, -\, 1}{x^5}\, \right)^{\frac{3}{2}}\, x^6\, + \, C\)

But I'm not "seeing" how to get to that....

Eliz.

 

[galactus]

Stapel, I worked this integral by hand and then ran it through Maple. Maple confirmed my answer. Yet, Mathematica gives the answer you posted. They're equivalent, I believe.


One trick is to use the substitution \(\displaystyle \L\\x=\frac{1}{u}\)

and \(\displaystyle dx=\frac{-1}{u^{2}}du\)

\(\displaystyle \L\\-\int\sqrt{\frac{\frac{1}{u}-1}{(\frac{1}{u})^{5}}}\cdot{\underbrace{\frac{1}{u^{2}}du}_{\text{dx}}}\)


Do the algebra and you'll see it whittles down to:

\(\displaystyle \L\\-\int\sqrt{1-u}du\)

\(\displaystyle =\L\\\frac{2(1-u)^{\frac{3}{2}}}{3}\)

Resub u=1/x:

\(\displaystyle =\L\\\frac{2(1-\frac{1}{x})^{\frac{3}{2}}}{3}\)

\(\displaystyle =\L\\\frac{2(\frac{x-1}{x})^{\frac{3}{2}}}{3}+C\)

 

[stapel]

Stapel, I worked this integral by hand and then ran it through Maple. Maple confirmed my answer. Yet, Mathematica gives the answer you posted. They're equivalent, I believe.

Yes, the two expressions are equivalent:

. . . . .\(\displaystyle \L \frac{2}{3}\, \left(\,\frac{x\, -\, 1}{x^5}\, \right)^{\frac{3}{2}}\, x^6\, =\, \frac{2}{3}\, \left(\,\frac{x\, -\, 1}{x}\, \right)^{\frac{3}{2}}\,\left(\, \frac{1}{x^4}\, \right)^{\frac{3}{2}}\, x^6\)

. . . . . . . . . . . . . . . . . . . .\(\displaystyle \L =\, \frac{2}{3}\, \left(\,\frac{x\, -\, 1}{x}\, \right)^{\frac{3}{2}}\,\left(\, \frac{1}{x^6}\, \right)\, x^6\)

. . . . . . . . . . . . . . . . . . . .\(\displaystyle \L =\, \frac{2}{3}\, \left(\,\frac{x\, -\, 1}{x}\, \right)\)

One trick is to use the substitution \(\displaystyle \L\\x=\frac{1}{u}\)

I'd have never thought of that substitution. Thank you!

Eliz.

 

[soroban]

Hello, k3232x!

I found a fairly painless approach . . .

\(\displaystyle \L\int\sqrt{\frac{x\,-\,1}{x^5}}\,dx\)

We have: \(\displaystyle \L\:\int\sqrt{\frac{x\,-\,1}{x^4\cdot x}} \,dx \;= \;\int\frac{1}{x^2}\sqrt{\frac{x\,-\,1}{x}}\,dx\)

Let \(\displaystyle u \:=\:\sqrt{\frac{x\,-\,1}{x}}\;\;\Rightarrow\;\;u^2\:=\:\frac{x\,-\,1}{x}\;\;\Rightarrow\;\;u^2x \:=\:x\,-\,1\;\;\Rightarrow\;\;x\,-\,u^2x\:=\:1\)

. . \(\displaystyle x(1\,-\,u^2}\:=\:1\;\;\Rightarrow\;\;x\:=\:\frac{1}{1\,-\,u^2}\;\;\Rightarrow\;\;dx\:=\:\frac{2u}{(1\,-\,u^2)^2}\,du\)


Substitute: \(\displaystyle \L\:\int\frac{1}{\left(\frac{1}{1-u^2}\right)^2}\,\cdot\,u\,\cdot\,\frac{2u}{(1\,-\,u^2)^2}\,du \;=\;\int2u^2\,du \;=\;\frac{2}{3}u^3\,+\,C\)

Back-substitute: \(\displaystyle \L\:\frac{2}{3}\left(\sqrt{\frac{x\,-\,1}{x}}\right)^3\,+\,C\;=\;\frac{2}{3}\left(\frac{x\,-\,1}{x}\right)^{\frac{3}{2}}\,+\,C\)

 

[pka]

There is nothing new in this post. However, this may make it obvious why (1/u) is the ideal substitution.
\(\displaystyle \L \frac{{\sqrt {x - 1} }}{{\sqrt {x^5 } }} = \frac{{\sqrt {x - 1} }}{{x^2 \sqrt x }} = \frac{1}{{x^2 }}\sqrt {1 - \frac{1}{x}}\).