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

The integral of 1/(sqrt(x)+1) is 2(sqrt(x)+1-ln|sqrt(x)+1|)+C

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

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

2 (x + 1 - ln x + 1) + C

Solution steps

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

Apply u-substitution

: \int \frac{2 ( u - 1 )}{u} d u

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

Expand

\frac{u - 1}{u} : 1 - \frac{1}{u}

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

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

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

\int 1 d u = u

\int \frac{1}{u} d u = ln ∣ u ∣

= 2 (u - ln ∣ u ∣)

Substitute back

u = x + 1

= 2 (x + 1 - ln x + 1)

Add a constant to the solution

= 2 (x + 1 - ln x + 1) + C

x \to - 1 lim (\frac{ln ( x )}{x - 1})

x \to - 2 - lim ((3 x^{2}) - 6 x + 2)

\frac{d}{d x} (4 - 5 x)

D (2 x^{2} e^{x})

\frac{d}{d x} (cos (1 - x))

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