Question

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. $$\int x^{4} (x^2+x+1)^{1/2}dx$$

Answer 1

`"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)`