How do you integrate #int dx/sqrt(x^2+2x)# using trig substitutions?

Question

4231 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2018-03-29T13:02:31

The answer is #=ln(|sqrt(x^2+2x)+x+1|)+C#

Complete the square in the denominator

#x^2+2x=x^2+2x+1-1=(x+1)^2-1#

Let #x+1=secu#, #=>#, #dx=secutanudu#

Therefore, the integral is

#I=int(dx)/(sqrt(x^2+2x))=int(dx)/sqrt((x+1)^2-1)#

#=int(secutanu du)/(sqrt(tan^2u))#

#=int(secudu)#

#=int(secu(tanu+secu)du)/(tanu+secu)#

Let #v=tanu+secu#, #=>#, #dv=(secutanu+sec^2u)du#

Therefore,

#I=int(dv)/v#

#=lnv#

#=ln(tanu+secu)#

#=ln(|sqrt((x+1)^2-1)+x+1|)+C#

#=ln(|sqrt(x^2+2x)+x+1|)+C#