Integrate x/(x+1)^1/2 - Solving a Tricky Integration Problem

  • Thread starter Thread starter mcelgiraffe
  • Start date Start date
  • Tags Tags
    Integrate
mcelgiraffe
Messages
10
Reaction score
0
Hi,

I am trying to work a problem that seems to have me stumped.

∫x/√(x+1) dx


I have tried to look at it as a right triangle with:
hypotenouse = √(x+1)
sideA = 1
sideB = √x

So I have:
cot^2 ∅=x, dx=-2cot∅csc^2 ∅ d∅
csc∅=√(x+1)

Working through the problem I have
-2∫(cot^2 ∅/csc∅) * cot∅csc^2 ∅ d∅
-2∫cot^3 ∅ * csc∅ d∅
-2∫(cos^3 ∅/sin^3 ∅) * (1/sin∅) d∅
-2∫cos^3 ∅/sin^4 ∅ d∅

Trying to solve it from here using more identities just keeps getting messier and I don't seem to be making any progress.

So, my main question is "am I on the right track?" or "is there an easier way that I am overlooking?"

Thank You,

James
 
Physics news on Phys.org
There's a way easier route. Try the substitution u=x+1. x=u-1. The triangles aren't helping at all.
 
See Dick's post for doing it the right way.

For triangle method:
http://img403.imageshack.us/img403/3350/33956029eq1.jpg

I don't know how you made it that complicated.
 
Last edited by a moderator:
I am not sure how I made it that difficult except that I have been staring at this way too long today. Dicks method was much easier and greatly appreciated. Thanks to both of you.

James
 
Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...
Back
Top