# Rewriting confusing fractions

#### shamieh

##### Active member
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.

#### Prove It

##### Well-known member
MHB Math Helper
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.

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
Ahh! Thank you!

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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
Ahh! Thank you!

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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
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