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

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

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

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

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

Solution steps

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

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

Simplify

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

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

Add a constant to the solution

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

e^{\frac{( - x ^{2} )}{4}}

n = 1 \sum \infty \frac{1}{n} + \frac{1}{n ^{2}}

\int \frac{x ^{2} + 4}{x} d x

x \to \infty lim (ln (x))

\int \frac{1}{x 12 x + 49} d x

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