I have issues solving the following integral:
$$\int\sqrt{x^3+8}~dx$$
I tried substitution and integration by parts, but with no use. I'm guessing I have to use some trigonometric substitution.
Can anybody help solve this integral?
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- 4$\begingroup$ Circular functions won't cut it. You'll need elliptic integrals. $\endgroup$David H– David H2014-09-22 20:31:34 +00:00Commented Sep 22, 2014 at 20:31
- $\begingroup$ @DavidH You're probably right. I tried to input that in wolfram alpha. It looks messy, and there is F (elliptical integral of the first kind) $\endgroup$Ant– Ant2014-09-22 20:39:30 +00:00Commented Sep 22, 2014 at 20:39
- 1$\begingroup$ Gradshteyn & Rhyzhik includes the following related integral (3.139.7): $$\int_u^1\sqrt{1-x^3}\,dx=\frac{1}{5}\left\{\sqrt[4]{27}F(\beta,\sin 75^\circ)-2u\sqrt{1-u^3}\right\}$$ with $F(\psi,k)$ as the incomplete elliptic integral of the first kind. $\endgroup$Semiclassical– Semiclassical2014-09-23 01:52:20 +00:00Commented Sep 23, 2014 at 1:52
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1 Answer
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For any real number $x$:
When $|x|\leq2$ ,
$$\begin{array}\int\sqrt{x^3+8}\,dx &=\int2\sqrt2\sqrt{\dfrac{x^3}{8}+1}\,dx\\ &=\int\sum\limits_{n=0}^\infty\dfrac{2\sqrt2(-1)^n(2n)!x^{3n}}{8^n4^n(n!)^2(1-2n)}\,dx\\ &=\int\sum\limits_{n=0}^\infty\dfrac{2\sqrt2(-1)^n(2n)!x^{3n}}{32^n(n!)^2(1-2n)}\,dx\\ &=\sum\limits_{n=0}^\infty\dfrac{2\sqrt2(-1)^n(2n)!x^{3n+1}}{32^n(n!)^2(1-2n)(3n+1)}+C\end{array}$$
When $|x|\geq2$ ,
$$\begin{array}\int\sqrt{x^3+8}\,dx &=\int x^\frac{3}{2}\sqrt{1+\dfrac{8}{x^3}}\,dx\\ &=\int x^\frac{3}{2}\sum\limits_{n=0}^\infty\dfrac{(-1)^n(2n)!8^n}{4^n(n!)^2(1-2n)x^{3n}}\,dx\\ &=\int\sum\limits_{n=0}^\infty\dfrac{(-1)^n(2n)!2^nx^{\frac{3}{2}-3n}}{(n!)^2(1-2n)}dx\\ &=\sum\limits_{n=0}^\infty\dfrac{(-1)^n(2n)!2^nx^{\frac{5}{2}-3n}}{(n!)^2(1-2n)\left(\dfrac{5}{2}-3n\right)}+C\\ &=\sum\limits_{n=0}^\infty\dfrac{(-1)^n(2n)!2^{n+1}}{(n!)^2(2n-1)(6n-5)x^{3n-\frac{5}{2}}}+C\end{array}$$
