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

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

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

\int \frac{1}{v ^{2}} 1 + \frac{1}{v} d v

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

Solution steps

\int \frac{1}{v ^{2}} 1 + \frac{1}{v} d v

Apply u-substitution

: \int - 1 + u d u

= \int - 1 + u d u

Take the constant out:

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

= - \int 1 + u d u

Apply u-substitution

: \int w d w

= - \int w d w

Apply the Power Rule

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

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

Substitute back

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

Join

1 + \frac{1}{v} : \frac{v + 1}{v}

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

Add a constant to the solution

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

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