# Calc 1 integration substitution and Implicit diff!

• November 23rd 2008, 04:49 PM
diggidy123
Calc 1 integration substitution and Implicit diff!
I need some help with 2 and 8.

2) I see that U = sqrt(x-1) and then du=1/(2sqrt(x-1)) but how do I get an integer to sub in for xdx? Usually in our past problems we would have like xdx and u=x^2 so du = 2x and the fraction is 1/2. I guess I'm just not seeing it so some further good instruction would be nice.

8.I have the answer key that my prof gave me and I can't get the answer. I keep coming up with:

y=ln(x^2+y^2)
y'=1/(x^2+y^2) * 2x +2yy'
y'-2yy' = 2x/(x^2+y^2)

y'=2x/(x^2+y^2-2y)

I don't see how you get the -2y in the denominator..
• November 23rd 2008, 04:59 PM
skeeter
$\int x\sqrt{x-1} \, dx$

$u = x-1$

$x = u+1$

$du = dx$

$\int (u+1)\sqrt{u} \, du$

now integrate.

$y = \ln(x^2+y^2)$

$y' = \frac{2x + 2yy'}{x^2 + y^2}$

$x^2y' + y^2 y' = 2x + 2yy'$

$x^2y' + y^2y' - 2yy' = 2x$

$y'(x^2 + y^2 - 2y) = 2x$

$y' = \frac{2x}{x^2 + y^2 - 2y}$
• November 23rd 2008, 05:22 PM
diggidy123
So when you integrate 2 you get:

((u+1)^2)/2 * (2/3U^(3/2))

That seems really whacky and I'm guessing I'm not seeing where you're coming from. Another guy I asked said let U = sqrt(x-1) then solve the polynomial but I didn't understand that either. Sorry; I'm not this big of a dunce usually.
• November 23rd 2008, 07:13 PM
diggidy123
Sorry to bump but I really need to know how to do number 2. I have no idea how to do that and I get an incredibly hairy equation when I take the gen. integral in my calc.
• November 24th 2008, 05:14 AM
skeeter
$\int_0^1 (u+1)\sqrt{u} \, du$

distribute the $\sqrt{u}$ ...

$\int_0^1 u^{\frac{3}{2}} + u^{\frac{1}{2}} \, du$

now can you integrate?