What is the integral of 6/(x^{-1)+2} ?

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

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integral of 6/(x^{-1)+2}

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

\frac{3}{2} + 3 x - \frac{3}{2} ln ∣ 1 + 2 x ∣ + C

Solution steps

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

Take the constant out:

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

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

Simplify

\frac{1}{x ^{- 1} + 2} : \frac{x}{1 + 2 x}

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

Apply u-substitution

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

= 6 \cdot \int \frac{u - 1}{4 u} d u

Take the constant out:

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

= 6 \cdot \frac{1}{4} \cdot \int \frac{u - 1}{u} d u

Expand

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

= 6 \cdot \frac{1}{4} \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

= 6 \cdot \frac{1}{4} (\int 1 d u - \int \frac{1}{u} d u)

\int 1 d u = u

\int \frac{1}{u} d u = ln ∣ u ∣

= 6 \cdot \frac{1}{4} (u - ln ∣ u ∣)

Substitute back

u = 1 + 2 x

= 6 \cdot \frac{1}{4} (1 + 2 x - ln ∣ 1 + 2 x ∣)

Simplify

6 \cdot \frac{1}{4} (1 + 2 x - ln ∣ 1 + 2 x ∣) : \frac{3}{2} + 3 x - \frac{3}{2} ln ∣ 1 + 2 x ∣

= \frac{3}{2} + 3 x - \frac{3}{2} ln ∣ 1 + 2 x ∣

Add a constant to the solution

= \frac{3}{2} + 3 x - \frac{3}{2} ln ∣ 1 + 2 x ∣ + C

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