1
$\begingroup$

I started to solve this problem as one which could be decomposed using partial fractions, and I think that you can do it, but the problem is that you end up with an imaginary number in the formulation and my solution is not matching up to the known solution.

Here is my attempt via partial fractions:

$$ \int{ \frac{1}{(x-1)^{3/2}x^{1/2}} dx } $$

Let $$ \frac{1}{(x-1)^{3/2}x^{1/2}} = \frac{A}{\sqrt{x}} + \frac{B}{(x-1)^{3/2}} $$

$$ 1=A(x-1)^{3/2} + B\sqrt{x} $$

Checking for x = 1 and x = 0 we get that $$ B = 1 \hspace{2em}A = \frac{1}{i} $$

Now $$ \int{ \frac{1}{(x-1)^{3/2}x^{1/2}} dx } = \int{\frac{1}{i\sqrt{x}}} dx + \int{\frac{1}{(x-1)^{3/2}}}dx = -2i\sqrt{x} -\frac{2}{\sqrt{x-1} } + C$$

I am close, but the solution should be $$ -\frac{2x}{\sqrt{x(x-1)}} + C $$ and one can get there using trig identities, but I am trying to solve this in the way that is more intuitive for me by partial fractions. I feel like I must be missing something obvious here. I would prefer to not have a full solution offered, but a hint so that I can work through it.

$\endgroup$
4
$\begingroup$

Your error is that no constants $A,\,B$ exist satisfying your putative identity. Since fractional powers are involved, partial fractions won't work, except perhaps with a clever substitution. If you want to do this without trigonometry, I recommend $u=x^a(x-1)^b$, where the best values of $a,\,b$ become obvious if you try it. But partial fractions won't be the crux of the solution.

$\endgroup$
1
  • $\begingroup$ Darn, thank you. $\endgroup$ – GrayLiterature Sep 4 '20 at 14:33
2
$\begingroup$

Following is the most intuitive solution of this problem I can think of. \begin{align} \int\dfrac{1}{(x-1)^\frac32x^\frac12}dx&=\int\frac{1}{x^2\left(\dfrac{x-1}{x}\right)^\frac32}dx\\ &=\int\frac{du}{u^\frac32}&&\left(\text{Substituting }u=1-\frac1x\right)\\ &=-\frac{2}{\sqrt u}+C \end{align}

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.