# Single Variable Calculus, Chapter 5, 5.4, Section 5.4, Problem 30

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Find the integrals $\displaystyle \int^2_1 \left( x + \frac{1}{x} \right)^2 dx$
\begin{aligned} \int\left( x + \frac{1}{x} \right)^2 dx &= \int \left( x^2 + 2 + \frac{1}{x^2} \right) dx \\ \\ \int\left( x + \frac{1}{x} \right)^2 dx &= \int x^2 dx + \int 2 dx + \int x^{-2} dx\\ \\ \int\left( x + \frac{1}{x} \right)^2 dx &= \frac{x^{2+1}}{2+1} + 2 \left( \frac{x^{0+1}}{0+1} \right) + \frac{x^{-2+1}}{-2+1} + C\\ \\ \int\left( x + \frac{1}{x} \right)^2 dx &= \frac{x^3}{3} + 2x - \frac{x^{-1}}{1} + C\\ \\ \int\left( x + \frac{1}{x} \right)^2 dx &= \frac{x^3}{3} + x - \frac{1}{x} + C\\ \\ \int^2_1 \left( x + \frac{1}{x} \right)^2 dx &= \frac{(2)^3}{3} + 2(2) - \frac{1}{2} + C \left[ \frac{(1)^3}{3} + 2 (1) - \frac{1}{1} + C \right]\\ \\ \int^2_1 \left( x + \frac{1}{x} \right)^2 dx &= \frac{8}{3} + 4 - \frac{1}{2} + C - \frac{1}{3} - 2 + 1 - C\\ \\ \int^2_1 \left( x + \frac{1}{x} \right)^2 dx &= \frac{29}{6} \end{aligned}