How do you use substitution to integrate #x^2sqrt(x^(4)+5)#?

1 Answer
Jul 2, 2015

If I rewrite this as:

#x^2 sqrt((x^2)^2 + (sqrt5)^2) prop u sqrt(u^2 + a^2#

then you can do the trig substitution method of letting:
#x^2 = sqrt5tantheta#
#sqrt(x^4 + 5) = sqrt5sec^2theta#
#x = 5^(1/4)sqrttantheta#
#dx = 5^(1/4)*1/(2sqrttantheta)sec^2thetad theta = 5^(1/4)/2sec^2theta/(sqrttantheta)d theta#

#= int sqrt5tantheta sqrt5sec^2theta 5^(1/4)/2sec^2theta/(sqrttantheta)d theta#

#= 5^(5/4)/2int (tantheta)^(1/2) sec^4thetad theta#

#= 5^(5/4)/2int (tantheta)^(1/2) (tan^2theta + 1)sec^2thetad theta#

since #1+tan^2theta = sec^2theta#.

Then just do some u-substitution with #u = tantheta# and #du = sec^2thetad theta# and you'll just have some polynomials to work with, integrate that, substitute back in previous variables, add #+C#.

I think you can do it from there (I'm in a hurry. If someone wants to finish this, go ahead).