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

[tomwilliam]

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

 

[praharmitra]

Your answer is correct as is the books. ln a + ln b = ln (ab).

 

[HallsofIvy]

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')

 

[tomwilliam]

Thanks! Nice to get one right for a change...