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

Question

I have tried integration by parts but couldn't work it out because of multiple $x$ variables in the square root. $$\int x^{4} (x^2+x+1)^{1/2}dx $$

Answer 1

write the integral in the form $$\int x^4\sqrt{\left(x+\frac{1}{2}\right)^2+\frac{3}{4}}dx$$ and substitute $$u=x+\frac{1}{2}$$ and then we get $du=dx$ and the we get $$\int\left(u-\frac{1}{2}\right)^4\sqrt{u^2+\frac{3}{4}}du$$ and substitute now $$u=\frac{\sqrt{3}}{2}\tan(s)$$ with $$du=\frac{\sqrt{3}}{2}\sec^2(s)ds$$

Answer 2

HINT.-The result is not as concise as I would like to have. Put $\dfrac{\sqrt3u-1}{2}=x$ so you get $$\int x^{4} (x^2+x+1)^{1/2}dx=\int\left(\dfrac{\sqrt3u-1}{2}\right)^4\sqrt{u^2+1}\space du$$ You must solve $$\int(A_1u^4+A_2u^3+A_3u^2+A_4u+A_5)\sqrt{u^2+1}\space du$$ and it is easy to solve each integral summand.