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

The integral of 1/(6+2sqrt(x)) is 3+sqrt(x)-3ln|3+sqrt(x)|+C

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integral of 1/(6+2sqrt(x))

\int \frac{1}{6 + 2 x} d x

3 + x - 3 ln 3 + x + C

Solution steps

\int \frac{1}{6 + 2 x} d x

Apply u-substitution

= \int \frac{u}{3 + u} d u

Apply u-substitution

= \int \frac{v - 3}{v} d v

Expand

\frac{v - 3}{v} : 1 - \frac{3}{v}

= \int 1 - \frac{3}{v} d v

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 v - \int \frac{3}{v} d v

\int 1 d v = v

\int \frac{3}{v} d v = 3 ln ∣ v ∣

= v - 3 ln ∣ v ∣

Substitute back

= 3 + x - 3 ln 3 + x

Add a constant to the solution

= 3 + x - 3 ln 3 + x + C

What is the integral of 1/(6+2sqrt(x)) ?
The integral of 1/(6+2sqrt(x)) is 3+sqrt(x)-3ln|3+sqrt(x)|+C