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

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

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

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

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

Solution steps

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

Apply u-substitution

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

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

Take the constant out:

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

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

Expand

\frac{v - 1}{v} : 1 - \frac{1}{v}

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

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 v - \int \frac{1}{v} d v)

\int 1 d v = v

\int \frac{1}{v} d v = ln ∣ v ∣

= 2 (v - ln ∣ v ∣)

Substitute back

v = u + 1

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

Add a constant to the solution

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

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