Let I(x) = ∫ (x + 7)/(x) dx and I(9) = 12 + 7 _ e 7 . If I(1) = α +...

Given,

$I\left(x\right) = ∫ \sqrt{\dfrac{x + 7}{x}}dx$

Now let, $\dfrac{x + 7}{x} = {t}^{2} ⇒ - \dfrac{7}{{x}^{2}}dx = 2tdt$

$⇒ dx = \dfrac{- 14t}{{\left({t}^{2} - 1\right)}^{2}}dt$

So, $I\left(x\right) = - 14 ∫ \dfrac{{t}^{2}}{{\left({t}^{2} - 1\right)}^{2}}dt$

$⇒ I\left(x\right) = - 14 ∫ \dfrac{dt}{\left({t}^{2} + \dfrac{1}{{t}^{2}} - 2\right)}$

$⇒ I\left(x\right) = \dfrac{- 14}{2} ∫ \left[\dfrac{\left(1 - \dfrac{1}{{t}^{2}}\right)}{{\left(t + \dfrac{1}{t}\right)}^{2} - 4} + \dfrac{\left(1 + \dfrac{1}{{t}^{2}}\right)}{{\left(t - \dfrac{1}{t}\right)}^{2}}\right]dt$

$⇒ I\left(x\right) = - 7\left(\dfrac{1}{4}\ln \left|\dfrac{t + \dfrac{1}{t} - 2}{t + \dfrac{1}{t} + 2}\right| - \dfrac{1}{t - \dfrac{1}{t}}\right) + c$

Now when $x = 9, t = \dfrac{4}{3}$

$⇒ I\left(9\right) = 12 + 7 \times \ln 7 = \dfrac{- 7}{4}ln{\left(\dfrac{1}{7}\right)}^{2} + 7 \times \dfrac{12}{7} + c$

$⇒ c = \dfrac{7}{2}\ln 7$

Now when $x = 1, t = 2\sqrt{2}$

$⇒ I\left(1\right) = + \dfrac{7}{4}ln{\left(\dfrac{2\sqrt{2} + 1}{2\sqrt{2} - 1}\right)}^{2} + 7 \times \dfrac{2\sqrt{2}}{7} + \dfrac{7}{2}\ln 7$

$⇒ I\left(1\right) = \dfrac{7}{2}\ln \left(\dfrac{{\left(2\sqrt{2} + 1\right)}^{2}}{7}\right) + 2\sqrt{2} + \dfrac{7}{2}\ln 7$

$⇒ I\left(1\right) = 7\ln \left(2\sqrt{2} + 1\right) - \dfrac{7}{2}\ln 7 + 2\sqrt{2} + \dfrac{7}{2}\ln 7$

$⇒ \alpha = 2\sqrt{2} ⇒ {\alpha}^{4} = 64$