How do you integrate #x( (x^2) +1 )^ (1/2)#?

1 Answer
mason m
Aug 19, 2017

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

Explanation:

#intx(x^2+1)^(1/2)dx#

Let #u=x^2+1#. Differentiating this shows that #du=2xcolor(white).dx#.

We already have #xcolor(white).dx# in the integrand, so all we need to do is multiply the integrand by #2#. To balance this out, multiply the exterior of the integral by #1//2#.

#=1/2intunderbrace((x^2+1)^(1/2))_(u^(1/2))overbrace((2xcolor(white).dx))^(du)#

#=1/2intu^(1/2)du#

#=1/2u^(3/2)/(3/2)#

#=1/3u^(3/2)#

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

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