What is the integral of (sqrt(1/t-1)) 1/(t^2) ?

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

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integral of (sqrt(1/t-1)) 1/(t^2)

\int (\frac{1}{t} - 1) \frac{1}{t ^{2}} d t

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

Solution steps

\int \frac{1}{t} - 1 \frac{1}{t ^{2}} d t

Apply u-substitution

: \int - u - 1 d u

= \int - u - 1 d u

Take the constant out:

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

= - \int u - 1 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} (\frac{1}{t} - 1)^{\frac{3}{2}}

Join

\frac{1}{t} - 1 : \frac{1 - t}{t}

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

Add a constant to the solution

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

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