Question

Integral of dx/{(x+1)sqrt(x^2+4x+2)} is given as -arcsinh{(x+2)/(xsqrt3)}, but I am not getting it.?
The hint given is 'Put 1/(x+1)=t'

Using it, the expression reduces to Int. -1/sqrt(1+2t-t^2). How to proceed further to get the desired result?

Answer 1

Using Wolfram I do not get that answer

dx/{(x+1)sqrt(x^2+4x+2)}

copied and pasted into
http://www.wolframalpha.com/widgets/view.jsp?id=893912ad72b01acedd76b68aa3ffc4df

gives
tan^-1 [ x/sqrt (x^2+4x+2) ] + c

Answer 2

∫ -dt/√(1+2t-t^2)
= ∫ -dt/√(2-(1+t)^2)

Now recall that
∫ du/√(a^2-u^2) = arcsin(u/a)

So, that gives us

-arcsin (1+t)/√2
= -arcsin (x+2)/((x+1)√2)

I don't see where the √3 comes from, nor the arcsinh.

I suspect a typo in the problem or the answer. Feel free to double-check my work...