What is the integral of 1/(x^2)sqrt(9+1/x) ?

The integral of 1/(x^2)sqrt(9+1/x) is -2/3 ((9x+1)/x)^{3/2}+C

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integral of 1/(x^2)sqrt(9+1/x)

\int \frac{1}{x ^{2}} 9 + \frac{1}{x} d x

- \frac{2}{3} (\frac{9 x + 1}{x})^{\frac{3}{2}} + C

Solution steps

\int \frac{1}{x ^{2}} 9 + \frac{1}{x} d x

Apply u-substitution

= \int - 9 + u d u

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= - \int 9 + u d u

Apply u-substitution

= - \int v d v

Apply the Power Rule

= - \frac{2}{3} v^{\frac{3}{2}}

Substitute back

= - \frac{2}{3} (9 + \frac{1}{x})^{\frac{3}{2}}

Join

9 + \frac{1}{x} : \frac{9 x + 1}{x}

= - \frac{2}{3} (\frac{9 x + 1}{x})^{\frac{3}{2}}

Add a constant to the solution

= - \frac{2}{3} (\frac{9 x + 1}{x})^{\frac{3}{2}} + C

What is the integral of 1/(x^2)sqrt(9+1/x) ?
The integral of 1/(x^2)sqrt(9+1/x) is -2/3 ((9x+1)/x)^{3/2}+C