What is the integral of 1/(sqrt(z)(\sqrt{z)+1)^2} ?

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

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integral of 1/(sqrt(z)(\sqrt{z)+1)^2}

\int \frac{1}{z ( z + 1 ) ^{2}} d z

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

Solution steps

\int \frac{1}{z ( z + 1 ) ^{2}} d z

Apply u-substitution

: \int \frac{2}{u ^{2}} d u

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

Take the constant out:

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

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

Apply the Power Rule

: - \frac{1}{u}

= 2 (- \frac{1}{u})

Substitute back

u = z + 1

= 2 (- \frac{1}{z + 1})

Simplify

2 (- \frac{1}{z + 1}) : - \frac{2}{z + 1}

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

Add a constant to the solution

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

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