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

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

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

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

2 ln 1 + n + C

Solution steps

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

Apply u-substitution

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

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

Add a constant to the solution

= 2 ln 1 + n + C

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