Integration Problem Solution - S.P. Thomson's Calculus Made Easy

  • Thread starter Thread starter tomwilliam
  • Start date Start date
  • Tags Tags
    Integration
tomwilliam
Messages
141
Reaction score
2

Homework Statement


I'm trying to do this problem from S. P. Thomson's Calculus Made Easy
(see attached image file)


Homework Equations





The Attempt at a Solution


I've produce the answer here, but it seems to be different from the mathcad solution, which is:
ln(-1+x)+ln(1+x)-ln(1+x^2)

Could someone help me see if (and if so, how) my answer is equivalent (without the constant). If it's wrong, I'd appreciate a pointer as to why, and if there is a quicker way of tackling this problem than by using partial fractions, could you let me know?
Thanks in advance
 

Attachments

  • ex 004.jpg
    ex 004.jpg
    24.5 KB · Views: 388
Physics news on Phys.org
Your answer is correct as is the books. ln a + ln b = ln (ab).
 
In fact, both answers could be written as
ln\left(\frac{x^2-1}{x^2+1}\right)+ C
or even
ln\left(C'\frac{x^2-1}{x^2+1}\right)
where C= ln(C')
 
Thanks! Nice to get one right for a change...
 
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