It's Rainin' Integrals

$\begin{aligned} \int_0^1 \frac {dx}{\sqrt{-\ln x}} & = \int_0^1 \frac {dx}{\sqrt{\ln \frac 1x}} & \small \color{#3D99F6} \text{Let }u = \frac 1x \implies du = - \frac {dx}{x^2} \\ & = - \int_{\infty}^1 \frac {du}{u^2\sqrt{\ln u}} & \small \color{#3D99F6} \text{Let }t = \ln u \implies e^t \ dt = du \\ & = \int_0^{\infty} t^{\frac 12 - 1}e^{-t} dt \\ & = \Gamma \left(\frac 12\right) = \sqrt \pi & \small \color{#3D99F6} \text{where }\Gamma (\cdot) \text{ denotes the gamma function.} \end{aligned}$

$\implies A = 1$

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$\begin{aligned} \int_0^1 \sqrt{-\ln x} \ dx & = \int_0^1 \sqrt{\ln \frac 1x} \ dx & \small \color{#3D99F6} \text{Let }u = \frac 1x \implies du = - \frac {dx}{x^2} \\ & = - \int_{\infty}^1 \frac {\sqrt{\ln u}}{u^2}du & \small \color{#3D99F6} \text{Let }t = \ln u \implies e^t \ dt = du \\ & = \int_0^{\infty} t^{\frac 32 - 1}e^{-t} dt \\ & = \Gamma \left(\frac 32\right) = \frac 12 \Gamma \left(\frac 12\right) = \frac {\sqrt \pi}2 \end{aligned}$

$\implies B = 2$

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$\begin{aligned} \int_0^1 \frac {\ln^2 x}{1+x^2} \ dx & = \int_0^1 \frac {\ln^2 x}{(x-i)(x+i)} \ dx & \small \color{#3D99F6} \text{By partial fraction} \\ & = \int_0^1 \left(\frac {i\ln^2 x}{2(x+i)} - \frac {i\ln^2 x}{2(x-i)}\right) dx & \small \color{#3D99F6} \text{Let }u = x+i, \ v = x-i \\ & = \frac i2 \left(\int_i^{1+i} \frac {\ln^2 (u-i)}u du - \int_{-i}^{1-i} \frac {\ln^2 (v+i)}v dv \right) & \small \color{#3D99F6} \text{By integration by parts} \\ & = \frac i2 \bigg(\cancel{\ln(-iu)\ln^2(u-i)\bigg|_i^{1+i}}^0 - 2\int_i^{1+i} \frac {\ln(-iu)\ln (u-i)}{u-i} du \\ & \quad \quad - \cancel{\ln(iv)\ln^2(v+i)\bigg|_{-i}^{1-i}}^0 + 2\int_{-i}^{1-i} \frac {\ln (iv)\ln (v+i)}{v+i} dv \bigg) \\ & = i\left(- \int_i^{1+i} \frac {\ln(-iu)\ln (u-i)}{u-i} du + \int_{-i}^{1-i} \frac {\ln (iv)\ln (v+i)}{v+i} dv\right) & \small \color{#3D99F6} \text{By integration by parts} \\ & = i \bigg(\cancel{\ln(u-i) \text{Li}_2 (iu+1) \bigg|_i^{1+i}}^0 - \int_i^{1+i} \frac {\text{Li}_2 (iu+1)}{\ln(u-i)} du \\ & \quad \quad - \cancel{\ln(v+i) \text{Li}_2 (1-iv) \bigg|_{-i}^{1-i}}^0 + \int_{-i}^{1-i} \frac {\text{Li}_2 (1-iv)}{\ln(v+i)} dv \bigg) & \small \color{#3D99F6} \text{Li}_s (z) \text{ is polylogarithm function.} \\ & = i \left(- \int_i^{1+i} \frac {\text{Li}_2 (iu+1)}{\ln(u-i)} du + \int_{-i}^{1-i} \frac {\text{Li}_2 (1-iv)}{\ln(v+i)} dv \right) \\ & = i \left(-\text{Li}_3 (iu+1) \bigg|_i^{1+i} + \text{Li}_3 (1-iv) \bigg|_{-i}^{1-i} \right) \\ & = i \left(-\text{Li}_3 (i) + \text{Li}_3 (0) + \text{Li}_3 (-i) - \text{Li}_3 (0) \right) \\ & = \frac {\pi^3}{16} \end{aligned}$

$\implies C = 3$ and $D = 16$

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$\begin{aligned} \int_0^\infty \frac {\ln x}{1+x^2} \ dx & = \frac 12 \int_0^\infty \left(\frac {\ln x}{1+x^2} + \frac {\ln \frac 1x}{x^2\left(1+ \frac 1{x^2} \right)} \right) dx & \small \color{#3D99F6} \text{Using }\int_0^\infty f(x) \ dx = \int_0^\infty \frac {f\left(\frac 1x\right)}{x^2} dx \\ & = \frac 12 \int_0^\infty \left(\frac {\ln x}{1+x^2} - \frac {\ln x}{x^2+ 1} \right) dx \\ &= 0 \end{aligned}$

$\implies E = 0$

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$\color{#D61F06} \text{Solution too tedius to show}$

$\begin{aligned} \int_0^\infty \frac {\ln x}{1+x^3} dx & = \int_0^\infty \frac {\ln x}{(x+1)(x^2-x+1)} dx = - \frac 2{27}\pi^2 \end{aligned}$

$\implies F = 27$

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$\begin{aligned} \int_{-1}^1 \frac {\cos x}{e^\frac 1x +1} dx & = \frac 12 \int_{-1}^1 \left(\frac {\cos x}{e^\frac 1x +1} + \frac {\cos (-x)}{e^{-\frac 1x} +1} \right) dx & \small \color{#3D99F6} \text{Using }\int_a^b f(x) \ dx = \frac 12 \int_a^b f(a+b-x) \ dx \\ & = \int_{-1}^1 \left(\frac {\cos x}{e^\frac 1x +1} + \frac {e^\frac 1x \cos x}{1+ e^\frac 1x} \right) dx \\ & = \frac 12 \int_{-1}^1 \cos x \ dx = \frac 12 \sin x\bigg|_{-1}^1 \\ & = \sin 1 \end{aligned}$

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$\begin{aligned} \int_0^\infty \frac {\sin^3 x}{x^2} dx & = \frac 34 \int_0^\infty \frac {\sin x}{x^2}dx - \frac 14 \int_0^\infty \frac {\sin 3x}{x^2} dx & \small \color{#3D99F6} \text{Using }\sin 3\theta = 3\sin \theta - 4 \sin^3 \theta \\ & = - \frac {3\sin x}{4x} \ \bigg|_0^\infty + \frac 34 \int_0^\infty \frac {\cos x}x dx + \frac {\sin 3x}{4x} \ \bigg|_0^\infty - \frac 14 \int_0^\infty \frac {3 \cos 3x}x dx & \small \color{#3D99F6} \text{By integration by parts} \\ & = - \frac 34 + \lim_{x \to 0} \frac 34 \text{Ci}(x) + \frac 34 - \lim_{x \to 0} \frac 34 \text{Ci}(3x) \\ & = \frac 34 \lim_{x \to 0} (\text{Ci}(x) - \text{Ci}(3x)) \\ & = \frac 34 \ln 3 \end{aligned}$

$\implies G = 4$

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$\color{#D61F06} \text{Solution too tedius to show}$

$\begin{aligned} \int_0^\infty \frac {\sin^5 x}{x^5} dx & = \int_0^\infty \frac {\sin 5x-5\sin 3x+10\sin x}{16x^5} dx = \frac {115}{384}\pi\end{aligned}$

$\implies H = 115$ and $I = 384$

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$\color{#D61F06} \text{Solution too tedius to show}$

$\begin{aligned} \int_0^\infty \frac {\sin^6 x}{x^6} dx & = \int_0^\infty \frac {10-15\cos 2x+6 \cos 4x-\cos 6x}{32x^5} dx = \frac {11}{40}\pi\end{aligned}$

$\implies J = 11$ and $K = 40$

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$\implies A+B+C+D+E+F+G+H+I+J+K = 1+2+3+16+0+27+4+115+384+11+40=\boxed{603}$