Trying to integrate [[1 -x^3]^(1/2)]/(x^4)

[mathlover97]

Hey everyone,

I'm just having some trouble by trying to integrate this function : [[1 -x^3]^(1/2)]/(x^4). I know that I can put u= [1 -x^3]^(1/2) and then raise both parts to the square to get rid of the radical and do some manupulation to end with : (-2/3)[(u^2)/(1-u^2)^2] . After that I integrate with the partial fractions technique but the problem that I have that I don't find the good values of the numerators for each partial fractions.

My partial fractions are : A/(1-u) + B/[(1-u)^2] + C/(1+u) + D/(1-u^2) and I find A=-1/4, B=1/4, C=-1/4, D=1/4

If somebody can help me out I'd be so grateful thank you!

 

[ksdhart2]

Alright, so just so we're both on the same page, this was the integral as given to you in the problem statement, yes?

\(\displaystyle \displaystyle \int \:\frac{\sqrt{1-x^3}}{x^4}dx\)

Assuming that's correct and using your u-substitution, this is what I get:

\(\displaystyle \displaystyle u=\sqrt{1-x^3}\), \(\displaystyle \displaystyle du=-\frac{3x^2}{2\sqrt{1-x^3}} \: dx\), \(\displaystyle \displaystyle x=\sqrt[3]{1-u^2}\), and \(\displaystyle \displaystyle dx=-\frac{2u}{3\left(1-u^2\right)^{\frac{2}{3}}} \: du\)

\(\displaystyle \displaystyle \int \:\frac{u}{\left(1-u^2\right)^{\frac{4}{3}}}\cdot -\frac{2u}{3\left(1-u^2\right)^{\frac{2}{3}}}du=-\int \:\frac{2u^2}{3\left(1-u^2\right)^2}du\)

This appears to match what you got. I also agree with your partial fraction decomposition, except for one minor caveat. The denominator of the D term should be (1 + u)^2. So the integral then becomes:

\(\displaystyle \displaystyle -\int \:\frac{2u^2}{3\left(1-u^2\right)^2}du=-\int \: \frac{-\frac{1}{4}}{u+1}+ \frac{\frac{1}{4}}{\left(u+1\right)^2} \:+\frac{\frac{1}{4}}{u-1}+\frac{\frac{1}{4}}{\left(u-1\right)^2}\)

From here, I'm not sure what's giving you pause. Do you think you did something wrong because the numerators aren't integers? The numerators don't have to be integers at all, although you can easily make them that if you prefer. You can either factor out a 1/4 term of all the partial fractions and then out of the integral, or recall what you learned way back in those algebra days:

\(\displaystyle \displaystyle \frac{\left(\frac{a}{b}\right)}{c}=\frac{a}{bc}\)

 

[mathlover97]

Thank you very much for that quick reply kdshart2 and for taking you time to answer my problem!
For the denominator of the D term I just didn't enter the good one on the topic but on paper I had the good one haha! The thing that I saw that wasn't the same as yours is the A and B denominators.... I have for each one (1 -u) and [(1-u)^2] but you have
( u-1) and [(u-1)^2]... I don't understand how you ended with those. With the ones that you have I'm able to find the right numerators but with mine I'm not obviously.....

Thank you!​

 

[ksdhart2]

Oh, yeah, you can switch the order around and use either (1 - u) or (u - 1) and it won't affect the answer. To see why that works, we can look at the integrand:

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

Factoring out the 2/3 was a step I forgot in my initial working. Hopefully you noticed that was an error on my part, and corrected it yourself. Sorry about that. Continuing from here, we can look at just the denominator:

\(\displaystyle \displaystyle (1-u^2)^2\)

For any u, \(\displaystyle \displaystyle (1-u^2)=-(u^2-1)\) but because we're actually squaring that (and squaring always results in a positive answer), we can see that \(\displaystyle \displaystyle (1-u^2)^2=(u^2-1)^2\)

And then we can factor the denominator:

\(\displaystyle \displaystyle \frac{2}{3} \cdot \frac{u^2}{(1-u^2)^2}=\frac{2}{3}\cdot \frac{1}{\left(u+1\right)^2\left(u-1\right)^2}=\frac{2}{3}\cdot \left(\frac{A}{u+1}+\frac{B}{\left(u+1\right)^2}+ \frac{C}{u-1}+\frac{D}{\left(u-1\right)^2}\right)\)

From which you'll get the proper values. If you don't factor our the 2/3 initially, your values will be 1/6 and -1/6 respectively instead of 1/4 and -1/4.

 

[mathlover97]

Thank you very much for your help !