Tricky Integral

• December 19th 2010, 11:44 AM
Chaobunny
Tricky Integral
So I have $\displaystyle \int{\frac{\sqrt{x}}{x+1}\,dx}$

My initial attempts at u substitution failed because I couldn't see any clear derivatives of terms in the numerator or denominator.

I looked at Wolfram Alpha and it says to use $u = \sqrt{x}$ and $du=\frac{1}{2\sqrt{x}}$

But then, I'm not really sure where to go from there. It says to go right to

$\displaystyle \2\int{\frac{u^2}{u^2+1}\,du}$

but I can't figure out how they got that from the u and du.
• December 19th 2010, 11:51 AM
Jhevon
Quote:

Originally Posted by Chaobunny
So I have $\displaystyle \int{\frac{\sqrt{x}}{x+1}\,dx}$

My initial attempts at u substitution failed because I couldn't see any clear derivatives of terms in the numerator or denominator.

I looked at Wolfram Alpha and it says to use $u = \sqrt{x}$ and $du=\frac{1}{2\sqrt{x}}$

But then, I'm not really sure where to go from there. It says to go right to

$\displaystyle \2\int{\frac{u^2}{u^2+1}\,du}$

but I can't figure out how they got that from the u and du.

one way (perhaps the easiest way):

$\displaystyle u = \sqrt x$

$\displaystyle \Rightarrow u^2 = x$

$\displaystyle \Rightarrow 2u ~du = dx$

$\displaystyle \int \frac {u}{u^2 + 1} \cdot 2 u~du$

$\displaystyle = 2 \int \frac {u^2}{u^2 + 1}~du$

otherwise, you'd do some voodoo like multiplying the original integral by $\frac {2 \sqrt x}{2 \sqrt x}$, and then make the switch from x to u

I assume you have no trouble taking it from there.
• December 19th 2010, 11:55 AM
Chaobunny
Ah, thank you very much! I never could quite get those u substitutions with the algebraic manipulations.
• December 19th 2010, 01:05 PM
Quote:

Originally Posted by Chaobunny
So I have $\displaystyle \int{\frac{\sqrt{x}}{x+1}\,dx}$

My initial attempts at u substitution failed because I couldn't see any clear derivatives of terms in the numerator or denominator.

I looked at Wolfram Alpha and it says to use $u = \sqrt{x}$ and $du=\frac{1}{2\sqrt{x}}$

It ought to have given

$\displaystyle\ u=\sqrt{x}\Rightarrow\frac{du}{dx}=\frac{1}{2\sqrt {x}}\Rightarrow\ dx=2\sqrt{x}du=2udu$

Then

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

Quote:

But then, I'm not really sure where to go from there. It says to go right to

$\displaystyle \2\int{\frac{u^2}{u^2+1}\,du}$

but I can't figure out how they got that from the u and du.
• December 19th 2010, 03:02 PM
Chaobunny
Oh, I forgot to add the 2 in the original post. But my main problem was with the manipulation of the du and dx, which was answered. So thank you both very much. =)