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

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

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

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

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

Solution steps

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

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 = x + 2

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

Simplify

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

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

Add a constant to the solution

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

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