<s>[INST] Definite integral involving square root of polynomial How can I solve the integral below?
$$\int_0^1 (x^4+x^3+x^2+x^1+1)^{1/2} \mathrm dx $$
 [/INST] You can use the horrible result of calculating $\int(x^4+x^3+x^2+x+1)^\frac{1}{2}~dx$ by Wolfram Integrator, but I have the following relatively simpler approach:
$\int_0^1(x^4+x^3+x^2+x+1)^\frac{1}{2}~dx$
$=\int_0^1\left(\dfrac{1-x^5}{1-x}\right)^{\frac{1}{2}}~dx$
$=\int_0^1\dfrac{(1-x^5)^\frac{1}{2}}{(1-x)^\frac{1}{2}}~dx$
$=\int_0^1\sum\limits_{n=0}^\infty\dfrac{(2n)!x^{5n}}{4^n(n!)^2(1-2n)(1-x)^\frac{1}{2}}dx$
$=\sum\limits_{n=0}^\infty\dfrac{(2n)!B\left(5n+1,\dfrac{1}{2}\right)}{4^n(n!)^2(1-2n)}$
$=\sum\limits_{n=0}^\infty\dfrac{(2n)!(5n)!\left(-\dfrac{1}{2}\right)!}{4^n(n!)^2\left(5n+\dfrac{1}{2}\right)!(1-2n)}$
$=\sum\limits_{n=0}^\infty\dfrac{(2n)!(5n)!\sqrt\pi}{4^n(n!)^2\dfrac{(10n)!\sqrt\pi}{4^{5n}(5n)!}(1-2n)}$
$=\sum\limits_{n=0}^\infty\dfrac{256^n(2n)!((5n)!)^2}{(n!)^2(10n)!(1-2n)}$
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