12
$\begingroup$

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?

$\endgroup$
  • 4
    $\begingroup$ Circular functions won't cut it. You'll need elliptic integrals. $\endgroup$ – David H Sep 22 '14 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 Sep 22 '14 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 Sep 23 '14 at 1:52
3
$\begingroup$

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 $|x|\geq2$ ,

$$\begin{array}\int\sqrt{x^3+8}\,dx &=\int x^\frac{3}{2}\sqrt{1+\dfrac{8}{x^3}}\,dx\\ &=\int x^\frac{3}{2}\sum\limits_{n=0}^\infty\dfrac{(-1)^n(2n)!8^n}{4^n(n!)^2(1-2n)x^{3n}}\,dx\\ &=\int\sum\limits_{n=0}^\infty\dfrac{(-1)^n(2n)!2^nx^{\frac{3}{2}-3n}}{(n!)^2(1-2n)}dx\\ &=\sum\limits_{n=0}^\infty\dfrac{(-1)^n(2n)!2^nx^{\frac{5}{2}-3n}}{(n!)^2(1-2n)\left(\dfrac{5}{2}-3n\right)}+C\\ &=\sum\limits_{n=0}^\infty\dfrac{(-1)^n(2n)!2^{n+1}}{(n!)^2(2n-1)(6n-5)x^{3n-\frac{5}{2}}}+C\end{array}$$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.