Given function is f(x)=(x+1x)^2 ⇒ f(x)=x^2+1x^2+2=x^2+x^ 2+2 So, ∫(x+1x)^2dx=∫(x^2+x^ 2+2)dx ⇒ ∫x^2 dx+∫x^ 2 dx+∫2 dx ⇒ x^33 1x+2x+c ...

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Antiderivative

Evaluate ∫x+1...

Question

Evaluate $\int {(x + \frac{1}{x})}^{2} d x$

\frac{x^{3}}{3} - \frac{1}{x} + x + c

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\frac{x^{3}}{3} + \frac{1}{x} + 2 x + c

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\frac{x^{3}}{3} - \frac{1}{x} + 2 x + c

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\frac{x^{3}}{3} + \frac{1}{x} + x + c

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Solution

The correct option is C $\frac{x^{3}}{3} - \frac{1}{x} + 2 x + c$
Given function is $f (x) = {(x + \frac{1}{x})}^{2}$
$\Rightarrow f (x) = x^{2} + \frac{1}{x^{2}} + 2 = x^{2} + x^{- 2} + 2$
So, $\int {(x + \frac{1}{x})}^{2} d x = \int (x^{2} + x^{- 2} + 2) d x$
$\Rightarrow \int x^{2} d x + \int x^{- 2} d x + \int 2 d x$
$\Rightarrow \frac{x^{3}}{3} - \frac{1}{x} + 2 x + c$

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