What is the integral of (sqrt(x)+1/x)^2 ?

The integral of (sqrt(x)+1/x)^2 is -1/x+4sqrt(x)+(x^2)/2+C

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integral of (sqrt(x)+1/x)^2

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

- \frac{1}{x} + 4 x + \frac{x ^{2}}{2} + C

Solution steps

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

Expand

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

= \int \frac{1}{x ^{2}} + \frac{2}{x ^{\frac{1}{2}}} + x 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 \frac{1}{x ^{2}} d x + \int \frac{2}{x ^{\frac{1}{2}}} d x + \int x d x

\int \frac{1}{x ^{2}} d x = - \frac{1}{x}

\int \frac{2}{x ^{\frac{1}{2}}} d x = 4 x^{\frac{1}{2}}

\int x d x = \frac{x ^{2}}{2}

= - \frac{1}{x} + 4 x^{\frac{1}{2}} + \frac{x ^{2}}{2}

Apply radical rule:

= - \frac{1}{x} + 4 x + \frac{x ^{2}}{2}

Add a constant to the solution

= - \frac{1}{x} + 4 x + \frac{x ^{2}}{2} + C

What is the integral of (sqrt(x)+1/x)^2 ?
The integral of (sqrt(x)+1/x)^2 is -1/x+4sqrt(x)+(x^2)/2+C