What is the integral of 1/(sqrt(x+9)) ?

The integral of 1/(sqrt(x+9)) is 2sqrt(x+9)+C

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integral of 1/(sqrt(x+9))

\int \frac{1}{x + 9} d x

2 x + 9 + C

Solution steps

\int \frac{1}{x + 9} 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 = x + 9

= 2 x + 9

Add a constant to the solution

= 2 x + 9 + C

\int \frac{4}{x ^{2} + 5 x - 14} d x

\int ln (14 x) d x

x \to 1 lim (cos (π x))

\int \frac{cos ( 2 x )}{sin ( 2 x )} d x

x \to 8 lim (\frac{x ^{2}}{4} - \frac{2}{x})

What is the integral of 1/(sqrt(x+9)) ?
The integral of 1/(sqrt(x+9)) is 2sqrt(x+9)+C