Integrate? Great!

\begin{aligned} I & = \int_0^1 \frac{1}{x^2+2x\cos \alpha +1} dx \\ & = \int_0^1 \frac{1}{x^2+2x\cos \alpha + \cos^2 \alpha +1 - \cos^2 \alpha} dx \\ & = \int_0^1 \frac{1}{(x +\cos \alpha)^2 + \sin^2\alpha } dx \\ & = \frac{1}{\sin^2 \alpha} \int_0^1 \frac{1}{\left( \color{#3D99F6} {\frac{x +\cos \alpha}{\sin \alpha}}\right)^2 + 1} dx \quad \quad \small \color{#3D99F6}{\text{Let }\tan \theta = \frac{x +\cos \alpha}{\sin \alpha} \implies \sec^2 \theta \ d\theta = \frac{dx}{\sin \alpha}} \\ & = \frac{\sin \alpha}{\sin^2 \alpha} \int_\color{#3D99F6}{\frac{\pi}{2} - \alpha}^\color{#3D99F6}{\frac{\pi}{2} - \frac{\alpha}{2}} \frac{\sec^2 \theta}{\sec^2 \theta} d\theta \quad \quad \small \color{#3D99F6}{\text{See Note.}} \\ & = \frac{1}{\sin \alpha} \int_{\frac{\pi}{2} - \alpha}^{\frac{\pi}{2} - \frac{\alpha}{2}} \ d\theta \\ & = \frac{1}{\sin \alpha} \left(\frac{\pi}{2}- \frac{\alpha}{2} - \frac{\pi}{2} + \alpha \right) \\ & = \frac{\alpha}{2\sin \alpha} \end{aligned}

$\implies a + 2b + 3c = 1 + 2(2) + 3(1) = \boxed{8}$

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$\color{#3D99F6}{\text{Note:}}$

$\begin{aligned} x = 1 \implies \tan \theta & = \frac{1+\cos \alpha}{\sin \alpha} \\ & = \frac{1+2\cos^2 \frac{\alpha}{2}-1}{2\sin \frac{\alpha}{2}\cos \frac{\alpha}{2}} \\ & = \cot \frac{\alpha}{2} \\ \implies \theta & = \frac{\pi}{2} - \frac{\alpha}{2} \\ x = 0 \implies \tan \theta & = \frac{0+\cos \alpha}{\sin \alpha} \\ & = \cot \alpha \\ \implies \theta & = \frac{\pi}{2} - \alpha \end{aligned}$

Let $I = \large \int_0^1 \dfrac1{x^2+2x\cos\alpha+1}\, dx$

Now, $\cos^2\alpha+\sin^2\alpha = 1$ . So, $\large\begin{aligned}I &= \int_0^1 \dfrac1{x^2+2x\cos\alpha+\cos^2\alpha+\sin^2 \alpha}\,dx \\\large&= \int_0^1 \dfrac1{\left(x+\cos\alpha\right)^2+\sin^2 \alpha}\, dx \\ \large&= \dfrac 1{\sin\alpha}\arctan\left(\dfrac{x+\cos\alpha}{\sin\alpha}\right)_0^1\quad \small\color{#3D99F6}{\because\int\dfrac{dx}{(x+a)^2+b^2} = \dfrac1b\arctan\dfrac{x+a}b}\\ \large&=\dfrac1{\sin\alpha}\left(\arctan\left(\dfrac{1+\cos\alpha}{\sin\alpha}\right)-\arctan\left(\dfrac{\cos\alpha}{\sin\alpha}\right)\right)\\ \large&=\dfrac1{\sin\alpha}\left(\arctan\left(\cot \dfrac \alpha2\right)-\arctan\left( \cot \alpha\right)\right)\\ \large&=\dfrac1{\sin\alpha}\left(\dfrac\pi2-\dfrac\alpha2-\dfrac\pi2+\alpha\right)\\ \large&=\dfrac\alpha{2\sin\alpha}\end{aligned}$

this one is jee style .assume alpha as pi/2 so now integraand turns into 1/(1+x^2).on integrating we get arctanx .substituting limits we get pi/4. now arrange according to given for so on numerator we have pi/2 and in denominator we have [2 times sin(pi/2)]on comparing the coefficient we have a=1 b=2 c=1