How far can Rick go?

ANSWER:

P = $\pi$

Q = $\pi \ln{ \left(\dfrac{\sqrt{\alpha+1}+1}{2}\right)}$

R = 38/3

3R = 38

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SOLUTION:

P: Proof that: $\int_{0}^{\infty} \dfrac{lnt}{1+t^2}\mathrm{d}t = 0$

Q:

$R = \int_{0}^{8} e^{ \ln{ \left(\dfrac{\sqrt{\alpha+1}+1}{2}\right)} }\mathrm{d}\alpha$

$R = \int_{0}^{8} \left(\dfrac{\sqrt{\alpha+1}+1}{2}\right)\mathrm{d}\alpha$

$R = \dfrac{38}{3}$

$3R = 38$

$\begin{aligned} P & = \int_0^1 \sqrt{\frac x{1-x}} \ln \left(\frac x{1-x}\right) dx & \small \blue{\text{Let }x = \sin^2 u \implies dx = 2\sin u \cos u} \\ & = \int_0^\frac \pi 2 4 \sin^2 u \ln (\tan u) \ du & \small \blue{\text{By integration by parts}} \\ & = -4\cos u \sin u \ln (\tan u) \bigg|_0^\frac \pi 2 + 4\int_0^\infty \left(\cos^2 u \ln (\tan u) + \frac {\cos u \sin u \sec^2 u}{\tan u} \right) du \\ & = 0 + 4\int_0^\frac \pi 2 \left((1-\sin^2 u) \ln (\tan u) + 1 \right) du \\ & = \blue{4 \int_0^\frac \pi 2 \ln(\tan u)\ du} - 4 \int_0^\frac \pi 2 \sin^2 u \ln (\tan u) + 4\int_0^\frac \pi 2 1 \ du & \small \blue{\text{By reflection: }\int_a^b f(x)\ dx = \int_a^b f(a+b-x)\ dx} \\ & = \blue{2 \int_0^\frac \pi 2 (\ln(\tan u)+ \ln(\cot u) )\ du} - P + 2\pi \\ & = 2 \int_0^\frac \pi 2 (\cancel{\ln(\tan u)} - \cancel{\ln(\tan u)})\ du - P + 2\pi \\ & = \pi \end{aligned}$

$\begin{aligned} Q(\blue \alpha) & = \int_0^\frac \pi 2 \ln \left(1+\blue \alpha \cos^2 x \right) dx \\ \frac {\partial Q(\alpha)}{\partial \alpha} & = \int_0^\frac \pi 2 \frac {\cos^2 x}{1+ \alpha \cos^2 x} dx = \int_0^\frac \pi 2 \frac 1{\sec^2 x + \alpha} dx = \int_0^\frac \pi 2 \frac 1{\tan^2 x + 1 + \alpha} dx & \small \blue{\text{Let } t = \tan x \implies dt = \sec^2 x \ dx} \\ & = \int_0^\infty \frac 1{(t^2+1+\alpha)(t^2+1)} dt = \frac 1\alpha \int_0^\infty \left(\frac 1{t^2+1} - \frac 1{t^2+1+\alpha}\right) dt \\ & = \frac \pi {2(\alpha+1 + \sqrt{\alpha+1})} \end{aligned}$

$\begin{aligned} \implies Q(\alpha) & = \int \frac \pi {2(\alpha+1 + \sqrt{\alpha+1})} d \alpha = \int \frac \pi{v+1} dv & \small \blue{\text{Let } v^2 =\alpha+1 \implies 2v\ dv = d\alpha} \\ & = \pi \ln (\sqrt{\alpha+1}+1) + C & \small \blue{\text{Sinc }Q(0) = 0 \implies C = - \pi ln 2} \\ & = \pi \ln \left(\frac {\sqrt{\alpha+1}+1}2 \right) \end{aligned}$

$\begin{aligned} R & = \int_0^8 \exp \left(\frac {\pi\ln \left(\frac {\sqrt{\alpha+1}+1}2 \right)}\pi \right) d\alpha = \int_0^8 \frac {\sqrt{\alpha+1}+1}2 d\alpha \\ & = \frac {(\alpha+1)^\frac 32}3 + \frac \alpha 2 \bigg|_0^8 = \frac {38}3 \end{aligned}$

Therefore $3R = \boxed{38}$ .