What is the integral of 8/(6+2sqrt(x)) ?

The integral of 8/(6+2sqrt(x)) is 8(3+sqrt(x)-3ln|3+sqrt(x)|)+C

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integral of 8/(6+2sqrt(x))

\int \frac{8}{6 + 2 x} d x

8 (3 + x - 3 ln 3 + x) + C

Solution steps

\int \frac{8}{6 + 2 x} d x

Take the constant out:

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

= 8 \cdot \int \frac{1}{6 + 2 x} d x

Apply u-substitution

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

= 8 \cdot \int \frac{u}{3 + u} d u

Apply u-substitution

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

= 8 \cdot \int \frac{v - 3}{v} d v

Expand

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

= 8 \cdot \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

= 8 (\int 1 d v - \int \frac{3}{v} d v)

\int 1 d v = v

\int \frac{3}{v} d v = 3 ln ∣ v ∣

= 8 (v - 3 ln ∣ v ∣)

Substitute back

= 8 (3 + x - 3 ln 3 + x)

Add a constant to the solution

= 8 (3 + x - 3 ln 3 + x) + C

\frac{\partial}{\partial x} (x e^{- t} sin (θ))

x \to 1 lim (\frac{\frac{2}{x} - 2}{x - 1})

\frac{\partial}{\partial x} (e^{x} + e^{- x})

f (x) = x^{3} - 8 x + 2

\frac{d}{d x} (3 x^{2} + x + 4)

What is the integral of 8/(6+2sqrt(x)) ?
The integral of 8/(6+2sqrt(x)) is 8(3+sqrt(x)-3ln|3+sqrt(x)|)+C