How do you integrate #int [xsqrt(x^2 + 1)] dx #?

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Answer 1 · 2015-06-08T00:47:41

This is of the form:

#sqrt(x^2 + a^2)#

which means you can use the substitution:

#x = atantheta#

So, let:
#x = tantheta#
#dx = sec^2thetad theta#
#sqrt(x^2+1) = sqrt(tan^2theta+1) = sqrt(sec^2theta) = sectheta#

#int xsqrt(x^2+1)dx = int tanthetasecthetasec^2thetad theta#

#= int (sec^2theta)(secthetatantheta)d theta#

Notice how you can do a second substitution here, with u-substitution.

Let:
#u = sectheta#
#du = secthetatanthetad theta#

#= int u^2du = u^3/3 + C#

#= sec^3theta/3 + C#

Since #sectheta = sqrt(x^2 + 1)#:

#=> 1/3(x^2+1)^(3/2) + C#