Let I(x)= (x+7)/(x)dx and I(9)=12+7 _e7.

Question

Let $I(x)=\int \sqrt{\frac{x+7}{x}}dx$ and $I(9)=12+7{\log }_{e}7$. If $I(1)=\alpha +7{\log }_{e}(1+2\sqrt{2})$, then ${\alpha }^{4}$ is equal to \_\_\_\_\_\_\_\_\_.

Answer: 64

Accepted Answer

Given integral: $\int \sqrt{\frac{x+7}{x}}\, dx$

Let's make the substitution $x={t}^{2}$. Then, $dx=2t\, dt$.

Substituting these values, the integral becomes :

$\int 2\sqrt{{t}^{2}+7}\, dt$

Now, let's evaluate this integral :

$I(t)=2(\frac{t}{2}\sqrt{{t}^{2}+7}+\frac{7}{2}\ln |t+\sqrt{{t}^{2}+7}|)+C$

Substituting back $t=\sqrt{x}$, we have :

$I(x)=2(\frac{\sqrt{x}}{2}\sqrt{x+7}+\frac{7}{2}\ln |\sqrt{x}+\sqrt{x+7}|)+C$

Simplifying further :

$I(x)=\sqrt{x}\sqrt{x+7}+7\ln |\sqrt{x}+\sqrt{x+7}|+C$
We are given that $I(9)=12+7\ln 7$.

Let's substitute $x=9$ and solve for the constant $C$:

$12+7\ln 7=\sqrt{9}\sqrt{9+7}+7\ln |\sqrt{9}+\sqrt{9+7}|+C$

$\Rightarrow$ $12+7\ln 7=3\sqrt{16}+7\ln |\sqrt{9}+\sqrt{16}|+C$

$\Rightarrow$ $12+7\ln 7=3\cdot 4+7\ln (3+\sqrt{16})+C$

$\Rightarrow$ $12+7\ln 7=12+7\ln (3+4)+C$

$\Rightarrow$ $12+7\ln 7=12+7\ln 7+C$

From this equation, we can see that $C=0$.

Now, we need to calculate $I(1)$ :

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

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

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

Therefore, $\alpha =\sqrt{8}$.

Finally, to find ${\alpha }^{4}$:

${\alpha }^{4}={(\sqrt{8})}^{4}$

$\Rightarrow$ ${\alpha }^{4}={8}^{2}$

$\Rightarrow$ ${\alpha }^{4}=64$

Hence, ${\alpha }^{4}$ is equal to 64.

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