Humongous power in the denominator!

Use reduction formula.

Let $I_{m,n} = \displaystyle \int_{0}^{\infty} (x^m+1)^{-n} dx$

Use, By parts, taking $dx$ as second function to get:

$I_{m,n} = \displaystyle \int_{0}^{\infty} \text{ m n }x^m (x^m +1)^{-(n+1)} dx$

$\Rightarrow I_{m,n} = \displaystyle \int_{0}^{\infty} \text{ m n } (1+x^m-1) (x^m+1)^{-(n+1)} dx$

$\Rightarrow I_{m,n} = \text{ m n } I_{m,n} - \text{ m n } I_{m,n+1}$

$\Rightarrow \boxed{I_{m,n+1} = \dfrac{\text{ m n-1 }}{\text{ m n }} I_{m,n}}$

Hence, we get $I_{3,5} = \dfrac{11}{12} \times \dfrac{8}{9} \times \dfrac{5}{6} \times \dfrac{2}{3} \times I_{3,1}$

$\Rightarrow I_{3,5} = \dfrac{110}{243} \displaystyle \int_{0}^{\infty} \dfrac{dx}{x^3+1} = 0.5474$

Note: You can find indefinite integral for $\int \frac{dx}{x^3+1}$ and put limits 0 to infity. I leave it on you.

Consider the integral:

$\displaystyle \int_0^{\infty} \frac{dx}{a^3+x^3}$

To evaluate the above, one can use partial fraction decomposition, however, I present a different approach. Use the substitution $x^3=a^3t$ to obtain:

$\displaystyle \int_0^{\infty} \frac{dx}{a^3+x^3}=\frac{1}{3a^2}\int_0^{\infty} \frac{t^{-2/3}}{1+t}\,dt=\frac{2\pi}{3\sqrt{3}a^2}$

where I used the result I proved in the following problem: Twelve? Ain't Nobody got time for that! Part 2

Differentiate wrt $a$ four times to get:

(I show each of the derivatives)

$\displaystyle \int_0^{\infty} \frac{3a^2}{(a^3+x^3)^2}\,dx=\frac{4\pi}{3\sqrt{3}a^3} \Rightarrow \int_0^{\infty}\frac{dx}{(a^3+x^3)^2}= \frac{4\pi}{9\sqrt{3}a^5}$

$\displaystyle \Rightarrow \int_0^{\infty}\frac{6a^2}{(a^3+x^3)^3}\,dx =\frac{20\pi}{9\sqrt{3}a^6} \Rightarrow \int_0^{\infty} \frac{dx}{(a^3+x^3)^3}=\frac{10\pi}{27\sqrt{3}a^8}$

$\displaystyle \Rightarrow \int_0^{\infty}\frac{9a^2}{(a^3+x^3)^4}\,dx=\frac{80\pi}{27\sqrt{3}a^9} \Rightarrow \int_0^{\infty}\frac{dx}{(a^3+x^3)^4}\,dx =\frac{80\pi}{243\sqrt{3}a^{11}}$

Finally differentiating once more,

$\displaystyle \Rightarrow \int_0^{\infty} \frac{12a^2}{(a^3+x^3)^5}=\frac{880\pi}{243\sqrt{3}a^{12}} \Rightarrow \int_0^{\infty} \frac{dx}{(a^3+x^3)^5}=\frac{220\pi}{729\sqrt{3}a^{14}}$

Substituting $a=1$ , hence, the final answer is:

$\int_0^{\infty} \frac{dx}{(1+x^3)^5}=\boxed{\dfrac{220\pi}{729\sqrt{3}}}$

So I used a contour integral around the path from 0 to $\infty$ , then around $120$ degrees, then back down to the origin. If $I$ is the integral, this gives $(1-\omega)I = 2\pi i \, {\rm Res}_{-w^2} \frac1{(z^3+1)^5}$ where $\omega = {\rm exp}(2\pi i/3)$ .

Computing the residue involves taking four derivatives of $\frac1{(z+1)^5(z+\omega)^5}$ and evaluating at $-\omega^2$ , so I'll cop to using some computer assistance there. At any rate, I got a residue of $\frac{110 \omega^2}{729}$ , which led to the same answer as everyone else: $\frac{220\pi\sqrt{3}}{2187}$ .

Substitute $x=t^{\frac{1}{3}}$ to get the answer as $\frac{1}{3}\Beta(\frac{1}{3},\frac{14}{3})$

We used the definition of beta function as $\large \Beta(m,n) = \int_{0}^{\infty} \frac{x^{n-1}}{(1+x)^{m+n}} dx$

Now using the following properties :-

$\Beta(m,n) = \frac{\Gamma(n)\Gamma(m)}{\Gamma(m+n)}$

$\Gamma(n+1) = n\Gamma(n)$

$\Gamma(n)\Gamma(1-n) = \frac{\pi}{\sin(\frac{\pi}{n})}$

We have our answer as

$\large\frac{11\cdot8\cdot5\cdot2}{3\cdot3\cdot3\cdot3\cdot3}\frac{\Gamma(\frac{1}{3})\Gamma(\frac{2}{3})}{\Gamma(5)}$

Evaluating this we get $\large \frac{660\pi}{3^{7}\sqrt{3}}$ as our answer.

Well, @Pranav Arora We can also do this using Beta Function.......... Simply substitute x raised to the power 1.5 as Tan(t) and then solve it........!!