What is the integral of 1/(x^2)sqrt(6+1/x) ?

The integral of 1/(x^2)sqrt(6+1/x) is -2/3 ((6x+1)/x)^{3/2}+C

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integral of 1/(x^2)sqrt(6+1/x)

\int \frac{1}{x ^{2}} 6 + \frac{1}{x} d x

- \frac{2}{3} (\frac{6 x + 1}{x})^{\frac{3}{2}} + C

Solution steps

\int \frac{1}{x ^{2}} 6 + \frac{1}{x} d x

Apply u-substitution

: \int - 6 + u d u

= \int - 6 + u d u

Take the constant out:

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

= - \int 6 + u d u

Apply u-substitution

: \int v d v

= - \int v d v

Apply the Power Rule

: \frac{2}{3} v^{\frac{3}{2}}

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

Substitute back

= - \frac{2}{3} (6 + \frac{1}{x})^{\frac{3}{2}}

Join

6 + \frac{1}{x} : \frac{6 x + 1}{x}

= - \frac{2}{3} (\frac{6 x + 1}{x})^{\frac{3}{2}}

Add a constant to the solution

= - \frac{2}{3} (\frac{6 x + 1}{x})^{\frac{3}{2}} + C

x \to 0 lim (cot (4 x))

\frac{d}{d x} (\frac{2 x ^{2} - 1}{x ^{3} - x})

\frac{1}{2 r} + \frac{1}{9 r ^{(\frac{8}{9})}}

x \to 0 lim (\frac{1}{3 x ^{3}})

y^{^{''}} - 2 y^{^{'}} + y = t^{3} e^{t}

What is the integral of 1/(x^2)sqrt(6+1/x) ?
The integral of 1/(x^2)sqrt(6+1/x) is -2/3 ((6x+1)/x)^{3/2}+C