Integral

**Mathstud28** · April 23rd 2008, 06:07 PM

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

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

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

**TheEmptySet** · April 23rd 2008, 06:19 PM

Originally Posted by Mathstud28

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

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

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

try $x=u^4 \to dx=4u^3du$

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

by long divison

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

from here it is off to the races

**Krizalid** · April 23rd 2008, 07:24 PM

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

**Jhevon** · April 23rd 2008, 07:36 PM

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

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

That wasn't so bad

...it was even easier knowing the EmptySet's final form to work towards...

**colby2152** · April 24th 2008, 04:43 AM

Originally Posted by Jhevon

That wasn't so bad

...it was even easier knowing the EmptySet's final form to work towards...

Of course!

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