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

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

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

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

2 ln x + 1 + C

Solution steps

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

Apply u-substitution

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

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

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= 2 ln ∣ u ∣

Substitute back

u = x + 1

= 2 ln x + 1

Add a constant to the solution

= 2 ln x + 1 + C

x^{3} - 3 x^{2} + 2 x - 2, at x = 2

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

\frac{d y}{d x} - \frac{2}{x} y = x

\int (e^{3.1})^{2} d x

\frac{d}{d x} (cos (2 sin (x)))

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