Hot Integrals

$I = \displaystyle \int_{0}^{1} \dfrac{1}{1+x^{3}}dx + \displaystyle \int_{1}^{\infty} \dfrac{1}{1+x^{3}}dx$
For the second integral, put $x = \dfrac{1}{t} \rightarrow dx = \dfrac{-dt}{t^{2}}$
$I = \displaystyle \int_{0}^{1} \dfrac{1}{1+x^{3}}dx +\displaystyle \int_{1}^{0} \dfrac{-t^{3}}{t^{2}(1+t^{3})}dt$ $I = \displaystyle \int_{0}^{1} \dfrac{1}{1+x^{3}}dx +\displaystyle \int_{0}^{1} \dfrac{x}{1+x^{3}}dx$
I used the properties
a) $\displaystyle \int_{a}^{b}f(t)dt = \displaystyle \int_{a}^{b}f(x) dx$
b) $\displaystyle \int_{a}^{b}f(x)dt = \displaystyle -\int_{b}^{a}f(x) dx$
$I = \displaystyle \int_{0}^{1}\dfrac{1+x}{1+x^{3}}dx$
$I = \displaystyle \int_{0}^{1} \dfrac{dx}{x^{2}-x+1}$
$\therefore I = \displaystyle \int_{0}^{1} \dfrac{dx}{\left(x-\dfrac{1}{2}\right)^{2} + \left(\dfrac{\sqrt{3}}{2}\right)^{2}}$
$\therefore I =\left[ \dfrac{2}{\sqrt{3}}\tan^{-1}\left(\dfrac{2t-1}{\sqrt{3}}\right) \right]_{0}^{1}$ $\therefore I = \dfrac{2}{\sqrt{3}} \cdot \dfrac{\pi}{3} = \dfrac{2\sqrt{3}\pi}{9}$
Comparing, A = 2, B = 3, C = 9
$A + B + C = 2 + 3 + 9 = 14$

Substitute $y=\dfrac{1}{1+x^3}$ . Note that as $x=0$ , $y=1$ and as $x\rightarrow \infty$ , $y=0$ . Also ,

$dy=\dfrac{-3x^2}{(x^3+1)^2} \ dx \text{(Using power rule)} \Rightarrow dx=-\dfrac{(x^3+1)^2dy}{3x^2} = -\dfrac{1}{3y}\left(\dfrac{1}{y} - 1\right)^{\frac{-2}{3}} \ dy$

Hence the integral becomes:

$\begin{aligned} \int_0^{\infty} \dfrac{dx}{(1+x^3)} &= \int_{1}^{0} -\dfrac{1}{3y}\left(\dfrac{1}{y} - 1\right)^{\frac{-2}{3}} \ dy \\ &= \dfrac{1}{3} \int_{0}^{1} \dfrac{y^{-1}(1-y)^{\frac{-2}{3}}}{y^{\frac{-2}{3}}} \ dy \\ &=\dfrac{1}{3} \int_{0}^{1} y^{-\frac{1}{3}}(1-y)^{-\frac{2}{3}} \ dy \\ &\text{Note that the above integral is in form of definition of Beta Function} \\ &= \dfrac{1}{3} B\left(\dfrac{2}{3} , \dfrac{1}{3}\right) \\ &=\dfrac{1}{3}\cdot\dfrac{\Gamma\left(\dfrac{2}{3}\right)\Gamma\left(\dfrac{1}{3}\right)}{\Gamma(1)} \\ &= \dfrac{1}{3} \cdot \Gamma\left(1-\dfrac{1}{3}\right)\Gamma\left(\dfrac{1}{3}\right) \quad\quad\quad \dots \ \because \Gamma(1)=1 \\ &=\dfrac{\pi}{3\sin\dfrac{\pi}{3}} \quad\quad\quad \dots \ \because \Gamma(n)\Gamma(1-n)=\dfrac{\pi}{\sin(n\pi)} \\ &=\dfrac{\pi}{3\left(\dfrac{\sqrt{3}}{2}\right)} \\ &=\dfrac{2\sqrt{3}\pi}{9} \end{aligned}$

Here, $A=2 \ , \ B=3 \ , \ C=9$ , hence , $A+B+C=2+3+9=\boxed{14}$