How to solve $$\int x^{4} (x^2+x+1)^{1/2}dx $$ ?

I have tried integration by parts but couldn't find a proper way to get rid of the square root. $$\int x^{4} (x^2+x+1)^{1/2}dx $$

1 Answer
MattyMatty
Mar 9, 2018

#"With the substitution"#

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

#"You can transform it immediately to an integral of a rational"#
#"function."#

#=> (t - x)^2 = t^2 - 2 t x + cancel(x^2) = cancel(x^2) + x + 1#
#=> x (1 + 2t) = t^2 - 1#
#=> x = (t^2 - 1)/(1 + 2t)#

#=> dx/{dt} = (2t(1+2t) - 2(t^2-1))/(1+2t)^2#
#= (2t^2+2t+2)/(1+2t)^2#
#= 2(t^2+t+1)/(1+2t)^2#

#x^2+x+1 = (t^2-1)^2/(1+2t)^2+(t^2-1)/(1+2t)+1#
#= (t^4 - 2 t^2 + 1 + (t^2-1)(1+2t) + (1+2t)^2)/(1+2t)^2#
#= (t^4 + 2 t^3 + 3 t^2 + 2 t + 1)/(1+2t)^2#
#= (t^2 + t + 1)^2/(1+2t)^2#
#=> sqrt(x^2+x+1) = (t^2+t+1)/(1+2t)#

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