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

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

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

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

2 ln x + 1 + 1 + C

Solution steps

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

Apply u-substitution

: \int \frac{2}{u + 1} d u

= \int \frac{2}{u + 1} 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 + 1} d u

Apply u-substitution

: \int \frac{1}{v} d v

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

Use the common integral:

\int \frac{1}{v} d v = ln (∣ v ∣)

= 2 ln ∣ v ∣

Substitute back

= 2 ln x + 1 + 1

Add a constant to the solution

= 2 ln x + 1 + 1 + C

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