Let $$I = \int\limits_6^{16} {{{{{\log }_e}{x^2}} \over {{{\log }_e}{x^2} + {{\log }_e}({x^2} - 44x + 484)}}dx} $$

$$I = \int\limits_6^{16} {{{{{\log }_e}{x^2}} \over {{{\log }_e}{x^2} + {{\log }_e}({x^2} - 22)}}dx} $$ .... (1)

We know,

$$\int\limits_a^b {f(x)dx} = \int\limits_a^b {f(a + b - x)} \,dx$$ (king)

So, $$I = \int\limits_6^{16} {{{{{\log }_e}{{(22 - x)}^2}} \over {{{\log }_e}{{(22 - x)}^2} + {{\log }_e}{{(21 - (22 - x))}^2}}}} $$

$$I = \int\limits_6^{16} {{{{{\log }_e}{{(22 - x)}^2}} \over {{{\log }_e}{x^2} + {{\log }_e}{{(22 - x)}^2}}}dx} $$ .... (2)

(1) + (2)

$$2I = \int\limits_6^{16} {1.\,dx} = 10$$

I = 5

$${U_n} = \prod\limits_{r = 1}^n {{{\left( {1 + {{{r^2}} \over {{n^2}}}} \right)}^r}} $$

$$L = \mathop {\lim }\limits_{n \to \infty } {({U_n})^{ - 4/{n^2}}}$$

$$\log L = \mathop {\lim }\limits_{n \to \infty } {{ - 4} \over {{n^2}}}\sum\limits_{r = 1}^n {\log {{\left( {1 + {{{r^2}} \over {{n^2}}}} \right)}^r}} $$

$$ \Rightarrow \log L = \mathop {\lim }\limits_{n \to \infty } \sum\limits_{r = 1}^n { - {{4r} \over n}.{1 \over n}\log \left( {1 + {{{r^2}} \over {{n^2}}}} \right)} $$

$$ \Rightarrow \log L = - 4\int\limits_0^1 {x\log (1 + {x^2})\,dx} $$

put 1 + x² = t

Now, 2xdx = dt

$$ = - 2\int\limits_1^2 {\log (t)dt = - 2[t\log t - t]_1^2} $$

$$ \Rightarrow \log L = - 2(2\log 2 - 1)$$

$$\therefore$$ $$L = {e^{ - 2(2\log 2 - 1)}}$$

$$ = {e^{ - 2\left( {\log \left( {{4 \over e}} \right)} \right)}}$$

$$ = {e^{\log {{\left( {{4 \over e}} \right)}^2}}}$$

$$ = {\left( {{e \over 4}} \right)^2} = {{{e^2}} \over {16}}$$

$$I = \int\limits_0^{{\pi \over 2}} {{{(1 + {{\sin }^2}x)} \over {(1 + {\pi ^{\sin x}})}} + {{{\pi ^{\sin x}}(1 + {{\sin }^2}x)} \over {(1 + {\pi ^{\sin x}})}}} dx$$

$$I = \int_0^{{\pi \over 2}} {(1 + {{\sin }^2}x)\,dx} $$

$$I = {\pi \over 2} + {\pi \over 2}.{1 \over 2} = {{3\pi } \over 4}$$

$$I = \int\limits_0^5 {{{x + [x]} \over {{e^{x - [x]}}}}dx = \alpha {e^{ - 1}} + \beta } $$

$$I = \int\limits_0^1 {{x \over {{e^x}}}dx + \int\limits_1^2 {{{x + 1} \over {{e^{x - 1}}}}dx + \int\limits_2^3 {{{x + 2} \over {{e^{x - 2}}}}dx + \int\limits_3^4 {{{x + 3} \over {{e^{x - 3}}}}dx + \int\limits_4^5 {{{x + 4} \over {{e^{x - 4}}}}dx} } } } } $$

Let $$I = {I_1} + {I_2} + {I_3} + {I_4} + {I_5}$$

Here, $${I_2} = \int\limits_1^2 {{{x + 1} \over {{e^{x - 1}}}}dx} $$ Put $$x = t + 1 \Rightarrow dx = dt$$

$$ = \int\limits_0^1 {{{t + 2} \over {{e^t}}}dt = \int\limits_0^1 {{t \over {{e^t}}}dt + \int\limits_0^1 {{2 \over {{e^t}}}dt} } } $$

$${I_2} = {I_1} + 2\int\limits_0^1 {{e^{ - t}}dt = {I_1} + 2(1 - {e^{ - 1}})} $$

Similarly,

$${I_3} = {I_1} + 4(1 - {e^{ - 1}})$$

$${I_4} = {I_1} + 6(1 - {e^{ - 1}})$$

$${I_5} = {I_1} + 8(1 - {e^{ - 1}})$$

$$I = {I_1} + {I_2} + {I_3} + {I_4} + {I_5} = 5{I_1} + (2 + 4 + 6 + 8)(1 - {e^{ - 1}})$$

$$ = 5{I_1} + 20(1 - {e^{ - 1}})$$

$${I_1} = \int_0^1 {x{e^{ - 1}}dx = - [{e^{ - x}}(x + 1)_0^1 = 1 - 2{e^{ - 1}}} $$

$$\therefore$$ $$5{I_1} + 20(1 - {e^{ - 1}}) = 5(1 - 2{e^{ - 1}}) + 20(1 - {e^{ - 1}}) = 25 - 30{e^{ - 1}}$$

$$\therefore$$ $$\alpha$$ = $$-$$30, $$\beta$$ = 25

Also it satisfy $$5\alpha + 6\beta = 0$$

Now, $${(\alpha + \beta )^2} = {( - 30 + 25)^2} = {( - 5)^2} = 25$$