M&S integration problem...

1. u= x+√(x^2+ a) ,
du/dx=[ 1+ x/√(x^2+a)]=[√(x^2+a) + x]/√(x^2+a)]
2. du = u/√(x^2+a) dx => dx/√(x^2+a) = du/u
=>∫ dx/√(x^2+a)
= ∫du/u = ln|u|+c
= ln| x+√(x^2+a) | + c

2009-01-07 19:13:15 補充：
There is a nice hint in your question.

could you please email me the photo for the question?
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