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Rewriting confusing fractions

shamieh

Active member
Sep 13, 2013
539
Confused on how we go from

\(\displaystyle \frac{1}{^4\sqrt{1 + x}}\) to \(\displaystyle \frac{4}{3}(1 + x)^\frac{3}{4}\)

Can someone please show me step-by-step. I need to see the basic steps.

Thanks in advance.
 

Prove It

Well-known member
MHB Math Helper
Jan 26, 2012
1,403
Confused on how we go from

\(\displaystyle \frac{1}{^4\sqrt{1 + x}}\) to \(\displaystyle \frac{4}{3}(1 + x)^\frac{3}{4}\)

Can someone please show me step-by-step. I need to see the basic steps.

Thanks in advance.
These things are NOT the same, so you can't "convert" them...

$\displaystyle \begin{align*} \frac{1}{\sqrt[4]{1 + x}} &= \frac{1}{ \left( 1 + x \right) ^{\frac{1}{4}} } \\ &= \left( 1 + x \right) ^{-\frac{1}{4}} \end{align*}$

It APPEARS though that you are trying to ANTIDIFFERENTIATE (Integrate) this function, which you should be able to do now...
 

shamieh

Active member
Sep 13, 2013
539
Ahh! Thank you!(Yes)

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And this particular problem would be divergent since you would get \(\displaystyle a^{3/4}\)which is > 1 correct?
 

Prove It

Well-known member
MHB Math Helper
Jan 26, 2012
1,403
Ahh! Thank you!(Yes)

- - - Updated - - -

And this particular problem would be divergent since you would get \(\displaystyle a^{3/4}\)which is > 1 correct?
What on EARTH are you talking about? WHAT is divergent? WHAT are you actually trying to do with this question?
 

shamieh

Active member
Sep 13, 2013
539
Oh sorry the original problem is the equation up top as \(\displaystyle \int^\infty_0\) and it's improper so i rewrote it as \(\displaystyle \int^a_0\) thus; \(\displaystyle \lim_{a\to\infty}\) and I ended up with a underneath the \(\displaystyle \sqrt{} \)to the \(\displaystyle ^3\) power.

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The initial question of the problem was Determine whether or not the integral is convergent or divergent. Sorry, forgot to include that.

So essentially I had this \(\displaystyle \lim_{a\to\infty} \frac{4}{3}(1 + a)^{3/4} - \frac{4}{3}\) so I'm guessing since it's \(\displaystyle \infty\) in the square root it's always going to keep growing no matter what and be Divergent