Let I ( x ) = x + 7 x d x and I ( 9 ) = 12 + 7 log e 7 . If I ( 1 ) = + 7 log e ( 1 + 2 2 ) then 4 is equal to _______ . [2023]

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Solution

Q.
Let $I (x) = \int \sqrt{\frac{x + 7}{x}} d x$ and $I (9) = 12 + 7 \log_{e} 7$ . If $I (1) = α + 7 \log_{e} (1 + 2 \sqrt{2})$ then $α^{4}$ is equal to _______ . [2023]

Ans.
(64)

$We have, I (x) = \int \sqrt{\frac{x + 7}{x}} d x$

$Put x = t^{2} \Rightarrow d x = 2 t d t$

$So, \int 2 \sqrt{t^{2} + 7} d t = 2 \int \sqrt{t^{2} + {(\sqrt{7})}^{2}} d t$

$= 2 [\frac{t}{2} \sqrt{t^{2} + 7} + \frac{7}{2} \ln | t + \sqrt{t^{2} + 7} |] + C$

$\Rightarrow I (x) = \sqrt{x} \sqrt{x + 7} + 7 \ln$ $| \sqrt{x} + \sqrt{x + 7} |$ $+ C$

$We have, I (9) = 12 + 7 \ln 7 = 12 + 7 [\ln (3 + 4)]$

$\Rightarrow C = 0$

$So, I (x) = \sqrt{x} \sqrt{x + 7} + 7 \ln$ $| \sqrt{x} + \sqrt{x + 7} |$

$I (1) = \sqrt{8} + 7 \ln$ $| 1 + \sqrt{8} |$

$∴$ $I (1) = \sqrt{8} + 7 \ln$ $| 1 + 2 \sqrt{2} |$ $\Rightarrow α = \sqrt{8}$

$∴$ $α^{4} = {[{(8)}^{1 / 2}]}^{4} = 8^{2} \Rightarrow α^{4} = 64$

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