What is the integral of 1/(sqrt(x)+4) ?

The integral of 1/(sqrt(x)+4) is 2(sqrt(x)+4-4ln|sqrt(x)+4|)+C

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integral of 1/(sqrt(x)+4)

\int \frac{1}{x + 4} d x

2 (x + 4 - 4 ln x + 4) + C

Solution steps

\int \frac{1}{x + 4} d x

Apply u-substitution

= \int \frac{2 ( u - 4 )}{u} d u

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= 2 \cdot \int \frac{u - 4}{u} d u

Expand

\frac{u - 4}{u} : 1 - \frac{4}{u}

= 2 \cdot \int 1 - \frac{4}{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

= 2 (\int 1 d u - \int \frac{4}{u} d u)

\int 1 d u = u

\int \frac{4}{u} d u = 4 ln ∣ u ∣

= 2 (u - 4 ln ∣ u ∣)

Substitute back

u = x + 4

= 2 (x + 4 - 4 ln x + 4)

Add a constant to the solution

= 2 (x + 4 - 4 ln x + 4) + C

What is the integral of 1/(sqrt(x)+4) ?
The integral of 1/(sqrt(x)+4) is 2(sqrt(x)+4-4ln|sqrt(x)+4|)+C