What is the integral of 1/(sqrt(6+x)) ?

The integral of 1/(sqrt(6+x)) is 2sqrt(6+x)+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of 1/(sqrt(6+x))

\int \frac{1}{6 + x} d x

2 6 + x + C

Solution steps

\int \frac{1}{6 + x} d x

Apply u-substitution

: \int 2 d u

= \int 2 d u

Integral of a constant:

\int a d x = a x

= 2 u

Substitute back

u = 6 + x

= 2 6 + x

Add a constant to the solution

= 2 6 + x + C

\int x^{3} arccot (x) d x

\int \frac{1}{( x + 2 ) ( x + 3 ) ( x + 4 )} d x

\int - cos (x) - sin (x) d x

\int (x^{2} + 4)^{10} x d x

What is the integral of 1/(sqrt(6+x)) ?
The integral of 1/(sqrt(6+x)) is 2sqrt(6+x)+C