Integration mania!

$\begin{aligned} I &=\displaystyle\int_{0}^{1}\dfrac{\log {\left(\frac{3t^2+4t+3}{3(t^2+1)}\right)}}{t}dt\\ &=\displaystyle\int_{0}^{1}\dfrac{\log{\left(1+\frac{2}{3}\frac{2t}{(1+t^2)}\right)}}{\frac{2t}{1+t^2}}\frac{2dt}{1+t^2} \end{aligned}$

$\text{Now doing the substitution} \ t=\tan{\left(\dfrac{x}{2}\right)}$ ,

$\text{The integral becomes}$ $I=\displaystyle\int_{0}^{\pi/2}\dfrac{\log {\left(1+\frac{2}{3}\sin x\right)}}{\sin x}dx$ .

$\text{Consider}$ $f(a)=\displaystyle\int_{0}^{\pi/2}\dfrac{\log ({1+\sin {a} \sin {x}})}{\sin x} dx \ , \ \ \ \ a \in \left[0,\frac{\pi}{2}\right]$ .

$\text{Using the}$ Leibniz integral rule $\text{and differentiating under the integral sign,}$ $\begin{aligned} f'(a)&=\displaystyle\int_{0}^{\pi/2} \dfrac{1}{\sin x} \dfrac{1}{1+\sin x \sin a}\cos a \sin x dx \\ &=\cos a\displaystyle\int_{0}^{\pi/2} \dfrac{dx}{1+\sin a \sin x} \end{aligned}$

$\text{Substituting} \ \tan {\frac{x}{2}}=t$ ,

$\begin{aligned} f'(a) &= \cos a \displaystyle\int_{0}^{1} \dfrac{\frac{2dt}{1+t^2}}{1+\sin a\frac{2t}{1+t^2}}\\ &=2\cos a \displaystyle\int_{0}^{1} \dfrac{dt}{(\sin a + t)^2+\cos^2 a}\\ &=2\cos a \dfrac{1}{|\cos a|} \left[\arctan {\left(\dfrac{t+\sin a}{|\cos a|}\right)}\right]_0^1 \\ \text{Now,} \ |\cos a|=\cos a, \ \because \ \ a \in \left[0,\dfrac{\pi}{2}\right]\\ \end{aligned}$ $\begin{aligned} \therefore f'(a) &=2\left(\arctan {\left(\dfrac{1+\sin a}{\cos a}\right)}-\arctan {(\tan a)}\right)\\ f'(a) &=2\left(\arctan {\left( \tan {\left(\dfrac{a}{2}+\dfrac{\pi}{4}\right)}\right)}+\arctan {(\tan a)}\right)\\ \text{So, because} \ \ a \in \left[0,\frac{\pi}{2}\right],\\ f'(a) &= 2\left(\dfrac{\pi}{4}-\dfrac{a}{2}\right)\\ \therefore f(a) &=\dfrac{1}{2}(\pi a-a^2)+C \end{aligned}$

$\text{We see, by plugging} \ a=0 \ \text{into} \ f(a), \ \text{that} \ C=0.$

$\therefore f(a) =\dfrac{1}{2}(\pi a-a^2)$

$\text{Now, all we need to do is find} \ I$ .

$\begin{aligned} I &= f\left(\arcsin {\dfrac{2}{3}}\right)\\ &=\dfrac{\arcsin {\dfrac{2}{3}}\left(\pi-\arcsin {\dfrac{2}{3}}\right)}{2}\\ \text{So,} \ \ \ \Large\color{#D61F06}{p+q+r+k=8} \end{aligned}$

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$\Huge\color{#3D99F6}{\mathbb{Quod \ Erat \ Demonstrandum}}$

Let the value of the required integral be $I$ . Now replacing $t$ by $\dfrac{1}{t}$ we get:

$I=\displaystyle\int_{1}^{\infty} \frac{\ln (3t^{2}+4t+3)-\ln (3t^{2}+3)}{t}\text{ }\text{d}t$

Thus $2I=\displaystyle\int_{0}^{\infty} \frac{\ln (3t^{2}+4t+3)-\ln (3t^{2}+3)}{t}\text{ }\text{d}t$ (As the integrand is continuous as at $t=1$ )

Now consider the integral

$\mathfrak{I}(k)=\displaystyle\int_{0}^{\infty} \frac{\ln (3t^{2}+kt+3)-\ln (3t^{2}+3)}{t}\text{ }\text{d}t$

Differentiating w.r.t. $k$ :

$\mathfrak{I}'(k)=\displaystyle\int_{0}^{\infty} \frac{1}{3t^{2}+kt+3}\text{ }\text{d}t$

$\therefore\mathfrak{I}'(k)=\dfrac{1}{3}\displaystyle\int_{0}^{\infty} \frac{1}{\left(t+\frac{k}{6}\right)^{2}+1-\frac{k^{2}}{36}}\text{ }\text{d}t$

As $k=4<6$ , $1-\dfrac{k^{2}}{36}>0$ .

Thus $\mathfrak{I}'(k)=\dfrac{2}{\sqrt{36-k^{2}}}\left(\dfrac{\pi}{2}-\arctan \left( \dfrac{k}{\sqrt{36-k^{2}}}\right)\right)$

Thus $\mathfrak{I}(4)-\mathfrak{I}(0)=2\bigg( \dfrac{\pi}{2}\arcsin \left(\dfrac{2}{3}\right)-\displaystyle\int_{0}^{4}\frac{1}{\sqrt{36-k^{2}}}\arcsin \left(\frac{k}{6}\right)\text{ }\text{d}k\bigg)$

Note that $\arctan \left( \dfrac{k}{\sqrt{36-k^{2}}}\right)=\arcsin \left(\dfrac{k}{6}\right)$ .Further note that $\mathfrak{I}(0)=0$ .

Now the integral above is easy to evaluate. Substitute $\arcsin \left(\dfrac{k}{6}\right)=y$ to obtain:

$\mathfrak{I}(4)=2\dfrac{\arcsin \left(\frac{2}{3}\right)\left(\pi-\arcsin \left(\frac{2}{3}\right)\right)}{2}$

Also note that $I=\dfrac{1}{2}\mathfrak{I}(4)$

Thus $I=\boxed{\dfrac{\arcsin \left(\frac{2}{3}\right)\left(\pi-\arcsin \left(\frac{2}{3}\right)\right)}{2}}$