What is the integral from 1 to 4 of (x+1/x)^2 ?

The integral from 1 to 4 of (x+1/x)^2 is 111/4

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integral from 1 to 4 of (x+1/x)^2

\int_{1}^{4} (x + \frac{1}{x})^{2} d x

\frac{111}{4}

Solution steps

\int_{1}^{4} (x + \frac{1}{x})^{2} d x

Expand

(x + \frac{1}{x})^{2} : x^{2} + 2 + \frac{1}{x ^{2}}

= \int_{1}^{4} x^{2} + 2 + \frac{1}{x ^{2}} d x

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= \int_{1}^{4} x^{2} d x + \int_{1}^{4} 2 d x + \int_{1}^{4} \frac{1}{x ^{2}} d x

\int_{1}^{4} x^{2} d x = 21

\int_{1}^{4} 2 d x = 6

\int_{1}^{4} \frac{1}{x ^{2}} d x = \frac{3}{4}

= 21 + 6 + \frac{3}{4}

Simplify

21 + 6 + \frac{3}{4} : \frac{111}{4}

= \frac{111}{4}

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What is the integral from 1 to 4 of (x+1/x)^2 ?
The integral from 1 to 4 of (x+1/x)^2 is 111/4