# Single Variable Calculus, Chapter 7, 7.4-1, Section 7.4-1, Problem 72

Find the integral $\displaystyle \int^9_4 \left( \sqrt{x} + \frac{1}{\sqrt{x}} \right)^2 dx$

\begin{aligned} \int^9_4 \left( \sqrt{x} + \frac{1}{\sqrt{x}}\right)^2 dx &= \int^9_4 \left( x + 2 \sqrt{x} \cdot \frac{1}{\sqrt{x}} + \frac{1}{x} \right) dx\\ \\ \int^9_4 \left( \sqrt{x} + \frac{1}{\sqrt{x}}\right)^2 dx &= \int^9_4 \left( x + 2 + \frac{1}{x} \right) dx\\ \\ \int^9_4 \left( \sqrt{x} + \frac{1}{\sqrt{x}}\right)^2 dx &= \left[ \frac{x^{1+1}}{1+1} + 2x + \ln x\right]^9_4\\ \\ \int^9_4 \left( \sqrt{x} + \frac{1}{\sqrt{x}}\right)^2 dx &= \left[ \frac{x^2}{2} + 2x + \ln x\right]^9_4\\ \\ \int^9_4 \left( \sqrt{x} + \frac{1}{\sqrt{x}}\right)^2 dx &= \frac{(9)^2}{2} + 2(9) + \ln 9 - \left[ \frac{(4)^2}{2} + 2(4) + \ln 4 \right]\\ \\ \int^9_4 \left( \sqrt{x} + \frac{1}{\sqrt{x}}\right)^2 dx &= \frac{81}{2} + 18 + \ln 9 - 8 - 8 - \ln 4\\ \\ \int^9_4 \left( \sqrt{x} + \frac{1}{\sqrt{x}}\right)^2 dx &= \frac{85}{2} + \ln 9 - \ln 4\\ \\ \int^9_4 \left( \sqrt{x} + \frac{1}{\sqrt{x}}\right)^2 dx &= \frac{85}{2} + \ln \frac{9}{4} \end{aligned}

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