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

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

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integral of 1/(9+sqrt(2x))

\int \frac{1}{9 + 2 x} d x

9 + 2 x - 9 ln 9 + 2 x + C

Solution steps

\int \frac{1}{9 + 2 x} d x

Apply u-substitution

= \int \frac{u - 9}{u} d u

Expand

\frac{u - 9}{u} : 1 - \frac{9}{u}

= \int 1 - \frac{9}{u} d u

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= \int 1 d u - \int \frac{9}{u} d u

\int 1 d u = u

\int \frac{9}{u} d u = 9 ln ∣ u ∣

= u - 9 ln ∣ u ∣

Substitute back

u = 9 + 2 x

= 9 + 2 x - 9 ln 9 + 2 x

Simplify

9 + 2 x - 9 ln 9 + 2 x : 9 + 2 x - 9 ln 9 + 2 x

= 9 + 2 x - 9 ln 9 + 2 x

Add a constant to the solution

= 9 + 2 x - 9 ln 9 + 2 x + C

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