How do you integrate #int x/sqrt(x^2+1)# by trigonometric substitution?

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Answer 1 · 2018-05-22T08:47:01

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

Let #x=tanu#

#=>#, #dx=sec^2udu#

#sqrt(x^2+1)=sqrt(1+tan^2u)=secu#

The integral is

#I=int(xdx)/sqrt(x^2+1)=int(tanusec^2udu)/(secu)#

#=inttanusecudu#

The derivative of #secu# is

#=(1/cosu)'=-1/cos^2u*-sinu=tanusecu#

Therefore,

The integral is

#I=secu=sec(arctanx)=sqrt(1+x^2)+C#