1. ## Integral

I dont know why but I am blanking on how to do this one

$\displaystyle \int\frac{x^{\frac{1}{4}}}{1+\sqrt{x}}dx$

I know its sad...it will be something obvious...I tam thinking trig sub?

2. Originally Posted by Mathstud28
I dont know why but I am blanking on how to do this one

$\displaystyle \int\frac{x^{\frac{1}{4}}}{1+\sqrt{x}}dx$

I know its sad...it will be something obvious...I tam thinking trig sub?
try $\displaystyle x=u^4 \to dx=4u^3du$

$\displaystyle \int\frac{x^{\frac{1}{4}}}{1+\sqrt{x}}dx=\int \frac{4u^4}{1+u^2}du$

by long divison

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

from here it is off to the races

3. To split the last integral, it's not necessary to do a long division. (Just in case.)

4. Originally Posted by Krizalid
To split the last integral, it's not necessary to do a long division. (Just in case.)
Let me see if I can tell what you are thinking

$\displaystyle \frac {4u^4}{1 + u^2} = \frac {4u^4 - 4 + 4}{1 + u^2} = \frac {4u^4 - 4}{1 + u^2} + \frac {4}{1 + u^2} =$ $\displaystyle \frac {4(u^2 + 1)(u^2 - 1)}{1 + u^2} + \frac 4{1 + u^2} = 4u^2 - 4 + \frac 4{1 + u^2}$

That wasn't so bad ...it was even easier knowing the EmptySet's final form to work towards...

5. Originally Posted by Jhevon
That wasn't so bad ...it was even easier knowing the EmptySet's final form to work towards...
Of course!