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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

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

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

= 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

\int (3 x - 1)^{6} d x

\int e^{7 x - 2} d x

\int \frac{cos ( x )}{5 - 3 cos ( x )} d x

\int tan (x) \cdot sin (x) d x

\int \frac{1 + x}{2} d x

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