What is the integral of 1/((x-1)^{3/2)} ?

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

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integral of 1/((x-1)^{3/2)}

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

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

Solution steps

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

Apply u-substitution

= \int \frac{1}{u ^{\frac{3}{2}}} d u

Apply the Power Rule

= - \frac{2}{u ^{\frac{1}{2}}}

Substitute back

u = x - 1

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

Add a constant to the solution

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

(\frac{y ^{^{'}} - e ^{- t} + 5}{y}) = - 5, y (0) = - 2

\int \frac{1}{k - x} d x

in v erse l a pl a ce \frac{3}{s ^{2} + 7}

\int 8 x d x

t an g e n t y = 16 x, (16, 64)

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