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Complete the square: $\int_0^1\frac{(3x^3-x^2+2x-4)}{\sqrt{(x-\frac{3}{2})^2}-\frac{1}{4}}\mathrm{d}x\\\\ =\int_0^12\frac{(3x^3-x^2+2x-4)}{\sqrt{(2x-3)^2}-1}\mathrm{d}x$ Now by substitution $u=2x-3$ such that

$\mathrm{d}x=\frac{1}{2} \mathrm{d} u \hspace{2cm} x^2=\frac{(u+3)^2}{4} \hspace{2cm} x^3=\frac{(u+3)^3}{8}\\\\ =\frac{1}{16}\int_0^1\frac{(3u^3+25u^2+77u+55)}{\sqrt{u^2-1}}\mathrm{d}u$

Now we have a simple trigonometric substitution $u=\sec(v)$

$\mathrm{d} u = \sec(v)\tan(v)\mathrm{d}v \\\\ \int_0^1\frac{\sec(v)[3u\sec^3(v)+25\sec^2(v)+77\sec(v)+55]\tan(v)}{\sqrt{\sec^2(v)-1}}\mathrm{d}v\\\\ =\int_0^1 \sec(v)[3u\sec^3(v)+25\sec^2(v)+77\sec(v)+55]\mathrm{d} v\\\\ =3\int_0^1 \sec^4(v)\mathrm{d} v +25\int_0^1 \sec^3(v)\mathrm{d} v+77\int_0^1 \sec^2(v)\mathrm{d}v +55\int_0^1 \sec(v)\mathrm{d} v$ The first integral:

$\int \sec^4(v)\mathrm{d}v \\\\ =\int \sec^2(v)[\tan^2(v)-1]\mathrm{d} v$ Let $w=\tan(v)$ , $\mathrm{d} w=\sec^2(v)\mathrm{d} v$

$\int (w^2 -1) \mathrm{d} w\\\\ = \frac{w^3}{3}-w\\\\ =\frac{\tan^3(v)}{3}-\tan(v)$

Then by the reduction formula we have that:

$\int \sin^n(v)\mathrm{d} v =\frac{n-2}{n-1}\int\sec^{n-2}(v)\mathrm{d} v +\frac{\sec^{n-2}(v)\tan (v)}{n-1}$ Therefore, for $n=3$

$\int\sec ^3(v)\mathrm{d} v = \frac{\sec (v)\tan (v)}{2}+\frac{1}{2}\int\sec (v)\mathrm{d}v$

We have a standard integral for n=2

$\int\sec ^2 (v)\mathrm{d} v = \tan (v)$ For $n=1$ , we expand and substitute: $\int\sec (v)\mathrm{d} v=\int\frac{\sec (v)[\tan (v)+\sec (v)]}{\tan (v)+\sec (v)}\mathrm{d} v$ Let

$u=\tan (v)+\sec (v) \hspace{2cm} \mathrm{d} v =\frac{1}{\sec^2(v)+\sec (v)\tan (v)}\mathrm{d} u\\\\ \int\sec (v)\mathrm{d} v = \int \frac{1}{u}\mathrm{d} u\\\\ =\ln u\\\\ =\ln (\tan (v)+\sec (v))$

Then finally we have:

$3\int_0^1 \sec^4(v)\mathrm{d} v +25\int_0^1 \sec^3(v)\mathrm{d} v+77\int_0^1 \sec^2(v)\mathrm{d}v +55\int_0^1 \sec(v)\mathrm{d} v\\\\ =\frac{135\ln (\tan (v) +\sec (v))}{2}+\tan^3(v)+\frac{25\tan (v)\sec (v)}{2}+80\tan (v)$

Now undo the substitutions one step at a time, simplify and we arrive at $\Big[\frac{135\ln (|2\sqrt{x^2-3x+2}+2x-3|)+\sqrt{x^2-3x+2}(16x^2+52x+202)}{16}\Big]_0^1=-2.981266944$

Therefore the PIN is 2981