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I was given the below definite integral to evaluate. below is the equation and my answer after I plug in the values but it is wrong. correct answer should be 29/6 what am I doing wrong? thank you

$$\int_1^2 \left(x + \frac 1x\right)^2 \,dx = \int_1^2 x^2 + \left(\frac 1x\right)^2 = \frac {x^3}{3} + \ln |x| = 3.0$$

(Source http://i.stack.imgur.com/V3UDP.jpg)

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    $\begingroup$ You should know that, $(a+b)^2=a^2+b^2+2ab$. :) $\endgroup$ – user112535 Dec 7 '13 at 18:15
  • $\begingroup$ oh you are right! dumbass me $\endgroup$ – Cash Vai Dec 7 '13 at 18:16
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You expanded incorrectly: Recall

$$(a + b)^2 = a^2 + 2ab + b^2$$

That gives us

$$\left(x + \frac 1x\right)^2 = x^2 + 2\left(x \cdot \frac 1x\right) + \frac 1{x^2} = x^2 + 2 + \frac 1{x^2}$$

Now go ahead and apply the power rule to evaluate the integral.

$$\begin{align} \int_1^2 \left(x + \frac 1x\right)^2 \,dx & = \int_1^2 \left( x^2 + 2 + \frac 1{x^2}\right)\,dx \\ \\ & = \int_1^2 \left( x^2 + 2 + x^{-2}\right)\,dx \\ \\ &= \dfrac {x^3}{3} + 2x - \frac 1x {\Huge|}_1^2\end{align}$$

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