Can anyone explain to me how to find the integral ?
$$ \int_0^1\sqrt{9x^4+4x^2+1}dx =? $$
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$\begingroup$A text:
Harris Hancock, Lectures on the Theory of Elliptic Functions, 498 pages (1909)
I have the Dover edition, 1958
I looked in Amazon ... it seems the soft cover edition is now selling for around \$10, and the Kindle edition is \$1 ...
$\endgroup$ 1 $\begingroup$$$\underbrace{\int\sqrt{9x^4+4x^2+1}\ dx}_J=x\sqrt{9x^4+4x^2+1}-\int x\frac{\big(9x^4+4x^2+1\big)'}{2\sqrt{9x^4+4x^2+1}}dx=x\sqrt{f(x)}-I.$$
$$x\cdot f'(x)=4\Big[f(x)-(2x^2+1)\Big]\iff I=2\ J-4\underbrace{\int\frac{x^2}{\sqrt{f(x)}}dx}_{I_2}-2\underbrace{\int\frac{dx}{\sqrt{f(x)}}}_{I_2}$$
$J=\dfrac{x\sqrt{f(x)}+4\ I_1+2\ I_2}3$ . For both integrals, let first $t=x^2$, then complete the square in their denominator, and use the fact that $\text{arcsin[h]}'u=\dfrac1{\sqrt{1\mp u^2}}$, with $u=u(t)$, in order to finally be able to express them in terms of elliptic integrals of $y(u)=w\big(\text{arcsin[h] }u\big)$.
$\endgroup$ 2 $\begingroup$Although this integral really belongs to an elliptic integral, but this does not means we can always express an elliptic integral to the elliptic integral of the three standard types conveniently, can tell us why.
Besides, it is better to tackle this integral to this series approach:
Hint:
$\int_0^1\sqrt{9x^4+4x^2+1}~dx$
$=\int_0^\frac{\sqrt{\sqrt{13}-2}}{3}\sqrt{9x^4+4x^2+1}~dx+\int_\frac{\sqrt{\sqrt{13}-2}}{3}^1\sqrt{9x^4+4x^2+1}~dx$ (separation according to the root between $0$ and $1$ of $9x^4+4x^2=1$)
$=\int_0^\frac{\sqrt{\sqrt{13}-2}}{3}\sqrt{1+x^2(9x^2+4)}~dx+\int_\frac{\sqrt{\sqrt{13}-2}}{3}^1x\sqrt{9x^2+4}\sqrt{1+\dfrac{1}{x^2(9x^2+4)}}~dx$
$=\int_0^\frac{\sqrt{\sqrt{13}-2}}{3}\sum\limits_{n=0}^\infty\dfrac{(-1)^n(2n)!(x^2(9x^2+4))^n}{4^n(n!)^2(1-2n)}dx+\int_\frac{\sqrt{\sqrt{13}-2}}{3}^1x\sqrt{9x^2+4}\sum\limits_{n=0}^\infty\dfrac{(-1)^n(2n)!}{4^n(n!)^2(1-2n)(x^2(9x^2+4))^n}dx$
$=\int_0^\frac{\sqrt{\sqrt{13}-2}}{3}\sum\limits_{n=0}^\infty\sum\limits_{k=0}^n\dfrac{(-1)^n(2n)!C_k^n4^{n-k}9^kx^{2n+2k}}{4^n(n!)^2(1-2n)}dx+\int_\frac{\sqrt{\sqrt{13}-2}}{3}^1x\sqrt{9x^2+4}\sum\limits_{n=0}^\infty\dfrac{(-1)^n(2n)!}{4^n(n!)^2(1-2n)x^{2n}(9x^2+4)^n}dx$
$=\int_0^\frac{\sqrt{\sqrt{13}-2}}{3}\sum\limits_{n=0}^\infty\sum\limits_{k=0}^n\dfrac{(-1)^n(2n)!9^kx^{2n+2k}}{4^kn!k!(n-k)!(1-2n)}dx+\int_\frac{\sqrt{\sqrt{13}-2}}{3}^1\sum\limits_{n=0}^\infty\dfrac{(-1)^n(2n)!}{4^n(n!)^2(1-2n)x^{2n-1}(9x^2+4)^{n-\frac{1}{2}}}dx$
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