What is the integral of (12)/(x^{-1)+3} ?

The integral of (12)/(x^{-1)+3} is 4/3+4x-4/3 ln|1+3x|+C

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integral of (12)/(x^{-1)+3}

\int \frac{12}{x ^{- 1} + 3} d x

\frac{4}{3} + 4 x - \frac{4}{3} ln ∣ 1 + 3 x ∣ + C

Solution steps

\int \frac{12}{x ^{- 1} + 3} d x

Take the constant out:

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

= 12 \cdot \int \frac{1}{x ^{- 1} + 3} d x

Simplify

\frac{1}{x ^{- 1} + 3} : \frac{x}{1 + 3 x}

= 12 \cdot \int \frac{x}{1 + 3 x} d x

Apply u-substitution

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

= 12 \cdot \int \frac{u - 1}{9 u} d u

Take the constant out:

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

= 12 \cdot \frac{1}{9} \cdot \int \frac{u - 1}{u} d u

Expand

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

= 12 \cdot \frac{1}{9} \cdot \int 1 - \frac{1}{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

= 12 \cdot \frac{1}{9} (\int 1 d u - \int \frac{1}{u} d u)

\int 1 d u = u

\int \frac{1}{u} d u = ln ∣ u ∣

= 12 \cdot \frac{1}{9} (u - ln ∣ u ∣)

Substitute back

u = 1 + 3 x

= 12 \cdot \frac{1}{9} (1 + 3 x - ln ∣ 1 + 3 x ∣)

Simplify

12 \cdot \frac{1}{9} (1 + 3 x - ln ∣ 1 + 3 x ∣) : \frac{4}{3} + 4 x - \frac{4}{3} ln ∣ 1 + 3 x ∣

= \frac{4}{3} + 4 x - \frac{4}{3} ln ∣ 1 + 3 x ∣

Add a constant to the solution

= \frac{4}{3} + 4 x - \frac{4}{3} ln ∣ 1 + 3 x ∣ + C

\frac{d y}{d x} = 3 + 12 x + 9 y + 36 x y

\frac{d}{d x} (\frac{1}{2 x + 9})

\int_{1}^{4} (2 x + 1) d x

\frac{\partial}{\partial x} (\frac{2 s - 3 x}{3 s + 5 x})

x \to - 2 lim (8 + x^{2})

What is the integral of (12)/(x^{-1)+3} ?
The integral of (12)/(x^{-1)+3} is 4/3+4x-4/3 ln|1+3x|+C