What is the integral of 1/(sqrt((x+1)^3)) ?

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

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integral of 1/(sqrt((x+1)^3))

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

- \frac{2}{x + 1} + C

Solution steps

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

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

assuming

(x + 1) \geq 0

= \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{2}{u}

Substitute back

u = x + 1

= - \frac{2}{x + 1}

Add a constant to the solution

= - \frac{2}{x + 1} + C

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