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

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

Apply the Power Rule

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

= - \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

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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