Prove the following

$=\int \frac{\frac{\sin x}{\cos ^{3} x}}{1+\tan ^{3} x} d x=\int \frac{\tan x \cdot \sec ^{2} x}{(\tan x+1)\left(1+\tan ^{2} x-\tan x\right)} d x$

Let $\tan x=t \Rightarrow \sec ^{2} x \cdot d x=d t$

$=\int \frac{\mathrm{t}}{(\mathrm{t}+1)\left(\mathrm{t}^{2}-\mathrm{t}+1\right)} \mathrm{dt}$

$=\int\left(\frac{\mathrm{A}}{\mathrm{t}+1}+\frac{\mathrm{B}(2 \mathrm{t}-1)}{\mathrm{t}^{2}-\mathrm{t}+1}+\frac{\mathrm{C}}{\mathrm{t}^{2}-\mathrm{t}+1}\right) \mathrm{dx}$

$\Rightarrow \mathrm{A}\left(\mathrm{t}^{2}-\mathrm{t}+1\right)+\mathrm{B}(2 \mathrm{t}-1)\left(\mathrm{t}^{2}-\mathrm{t}+1\right)+\mathrm{C}(\mathrm{t}+1)=\mathrm{t}$

$\Rightarrow \mathrm{t}^{2}(\mathrm{~A}+2 \mathrm{~B})+\mathrm{t}(-\mathrm{A}+\mathrm{B}+\mathrm{C})+\mathrm{A}-\mathrm{B}+\mathrm{C}=1$

$\therefore A+2 B=0$          .............(1)

$-\mathrm{A}+\mathrm{B}+\mathrm{C}=1$    ..........(2)

$\mathrm{~A}-\mathrm{B}+\mathrm{C}=0$    ............(3)

$\Rightarrow \mathrm{C}=\frac{1}{2} \Rightarrow \mathrm{A}-\mathrm{B}=-\frac{1}{2}$ .........(4)

$A+2 B=0$

$\mathrm{A}-\mathrm{B}=-\frac{1}{2}$

$\Rightarrow 3 \mathrm{~B}=\frac{1}{2} \Rightarrow \mathrm{B}=\frac{1}{6}$

$\mathrm{A}=-\frac{1}{3}$

$\mathrm{I}=-\frac{1}{3} \int \frac{\mathrm{dt}}{1+\mathrm{t}}+\frac{1}{6} \int \frac{2 \mathrm{t}-1}{\mathrm{t}^{2}-\mathrm{t}+1} \mathrm{dt}+\frac{1}{2} \int \frac{\mathrm{dt}}{\mathrm{t}^{2}-\mathrm{t}+1}$

$=-\frac{1}{3} \ell \mathrm{n}|(1+\tan \mathrm{x})|+\frac{1}{6} \ell \mathrm{n}\left|\tan ^{2} \mathrm{x}-\tan \mathrm{x}+1\right|$

$+\frac{1}{2} \cdot \frac{2}{\sqrt{3}} \tan ^{-1}\left(\frac{\left(\tan x-\frac{1}{2}\right)}{\frac{\sqrt{3}}{2}}\right)$

$=-\frac{1}{3} \ell \ln |(1+\tan x)|+\frac{1}{6} \ell \operatorname{n}\left|\tan ^{2} x-\tan x+1\right|$

$+\frac{1}{\sqrt{3}} \tan ^{-1}\left(\frac{2 \tan x-1}{\sqrt{3}}\right)+C$

$\alpha=-\frac{1}{3}, \beta=\frac{1}{6}, \gamma=\frac{1}{\sqrt{3}}$

$18\left(\alpha+\beta+\gamma^{2}\right)=18\left(-\frac{1}{3}+\frac{1}{6}+\frac{1}{3}\right)=3$