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

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

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

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

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

Solution steps

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

Take the constant out:

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

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

Simplify

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

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

Apply u-substitution

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

= 3 \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

= 3 \cdot \frac{1}{9} \cdot \int \frac{u - 1}{u} d u

Expand

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

= 3 \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

= 3 \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 ∣

= 3 \cdot \frac{1}{9} (u - ln ∣ u ∣)

Substitute back

u = 1 + 3 x

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

Simplify

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

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

Add a constant to the solution

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

\int 6 - x^{2} d x

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

2 x^{3} + 3 x^{2} - 12 x + 7

\frac{d}{d t} (\frac{t ^{3}}{3})

2 x^{2} + x + 1

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