What is the integral of 1/(3+sqrt(x)) ?

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

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integral of 1/(3+sqrt(x))

\int \frac{1}{3 + x} d x

2 (3 + x - 3 ln 3 + x) + C

Solution steps

\int \frac{1}{3 + x} d x

Apply u-substitution

: \int \frac{2 ( u - 3 )}{u} d u

= \int \frac{2 ( u - 3 )}{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 - 3}{u} d u

Expand

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

= 2 \cdot \int 1 - \frac{3}{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{3}{u} d u)

\int 1 d u = u

\int \frac{3}{u} d u = 3 ln ∣ u ∣

= 2 (u - 3 ln ∣ u ∣)

Substitute back

u = 3 + x

= 2 (3 + x - 3 ln 3 + x)

Add a constant to the solution

= 2 (3 + x - 3 ln 3 + x) + C

f (x) = ln (6 x^{3} + e^{x})

\frac{d}{d x} (\frac{1}{( x ^{2} - 3 x ) ^{2}})

y^{^{'}} = \frac{y - x}{x}

\frac{\partial}{\partial x} (4 x^{5} y)

\frac{\partial}{\partial x} (\frac{x ^{2} - 1}{x ^{2} + x + 1})

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