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________.
Step-by-Step Solution
Key Concept: Substitute x = 7 tan^2(theta) or use the substitution sqrt(x+7/x) = t to evaluate the integral.
Let $I = \int \sqrt{\frac{x+7}{x}} dx$. Let $\sqrt{x+7/x} = t$, then $t^2 = 1 + 7/x$, so $x = 7/(t^2-1)$. $dx = -14t/(t^2-1)^2 dt$. The integral becomes $\int t \cdot \frac{-14t}{(t^2-1)^2} dt = -14 \int \frac{t^2}{(t^2-1)^2} dt$. Using partial fractions or standard forms, we find $I(x) = \sqrt{x^2+7x} + \frac{7}{2} \ln|2x+7+2\sqrt{x^2+7x}| + C$. Using $I(9) = 12 + 7 \ln 7$, we find $C$. Then calculate $I(1)$ to find $\alpha = 4$. Thus $\alpha^4 = 64$.
Correct Answer: 64