# Thread: A Hard Calculus Problem

1. ## A Hard Calculus Problem

The variables t and x are related by t = x+√(x^2+2bx+c) where b and c are constants and
b^2<c. Show that

dx/dt=(t-x)/(t+b)

and hence integrate

1/√(x^2+2bx+c)

Verify by direct integration that your result holds also in the case b^2 = c if x+b > 0 but that
your result does not hold in the case b^2 = c if x + b < 0 .

[and come someone tell me how to use the math symbols at this site.. i had to do everything in microsoft word ]

2. Originally Posted by iPod
The variables t and x are related by t = x+√(x^2+2bx+c) where b and c are constants and

b^2<c. Show that

dx/dt=(t-x)/(t+b)

and hence integrate

1/√(x^2+2bx+c)

Verify by direct integration that your result holds also in the case b^2 = c if x+b > 0 but that
your result does not hold in the case b^2 = c if x + b < 0 .

[and come someone tell me how to use the math symbols at this site.. i had to do everything in microsoft word ]
Differentiate implicitly x with respect to t. This thread does not belong in this sub-forum.

3. I managed to differentiate, however Im just stuck at the intergration part,
and doesnt this thread belong here??

4. Hello iPod

Welcome to Math Help Forum!
Originally Posted by iPod
The variables t and x are related by t = x+√(x^2+2bx+c) where b and c are constants and
b^2<c. Show that

dx/dt=(t-x)/(t+b)
[CENTER]
$t = x+\sqrt{x^2+2bx+c}$

$\Rightarrow (t-x)^2 = x^2+2bx+c$

$\Rightarrow 2(t-x)\left(\frac{dt}{dx}-1\right)=2x+2b$

$\Rightarrow \frac{dt}{dx} = 1+\frac{x+b}{t-x}=\frac{t+b}{t-x}$

$\Rightarrow \frac{dx}{dt} =\frac{t-x}{t+b}$

and hence integrate

1/√(x^2+2bx+c)
$t-x = \sqrt{x^2+2bx+c}$

$\Rightarrow \frac{dx}{dt}=\frac{\sqrt{x^2+2bx+c}}{t+b}$

$\Rightarrow \int \frac{1}{\sqrt{x^2+2bx+c}}\,dx=\int\frac{1}{(t+b)} \,dt$

$= \ln (t+b)+C$

$=\ln(b+x+\sqrt{x^2+2bx+c})+C$, provided $b+x+\sqrt{x^2+2bx+c}>0$

Can you finish up?

PS
[and come someone tell me how to use the math symbols at this site.. i had to do everything in microsoft word ]
The language is called LaTeX, and you'll find some help on this site here: http://www.mathhelpforum.com/math-help/latex-help/

I also find the Wikipedia site helpful, here: LaTeX/Mathematics - Wikibooks, collection of open-content textbooks

When you've written some LaTex code, select what you've written and click the $\Sigma$ button on this editor's toolbar, and the editor will insert $$...$$ tags around the selection. Use 'Preview Changes' to check that it's OK.

To see any of the code I've used in my solution above, just click-left on any line and the code will appear in a separate pop-up window. Go on, try it now!

5. Oh yeah , it is known as Euler's substitution for integration .

We can solve integrals in this form $\int \frac{ P(x) }{\sqrt{x^2 + 2ax + b}}~dx ~~$ using Euler's substitution $t = x + \sqrt{x^2 + ax + b}$

$dt = ( 1 + \frac{x + a }{\sqrt{x^2 + 2ax + b}})~dx$

$dt = [ \frac{ ( \sqrt{x^2 + 2ax + b}+x) + a}{ \sqrt{x^2 + 2ax + b}}]~dx$

Therefore , the integral becomes

$\int \frac{ P( f(t)) }{ (t+a)}~dx~dt$

Then , what does f(t) refer to ?

Since

$t = x + \sqrt{x^2 + 2ax + b}$
$t^2 - 2tx = 2ax + b$

$x = \frac{t^2 -b}{2(a+t)}$

$x = f(t) = \frac{t^2 -b}{2(a+t)}$

For more details , you can take a look at http://planetmath.org/encyclopedia/E...tegration.html .

Euler's substitution is a powerful tools indeed . Make use of it !!

6. ## thx

Thank you very much Grandad and Simple, however I am not very familiar with this type of integration - is there a certain name for this type of integration and also is there a step by step method for it??