What is the integral of (2x)/((x+3)(3x+1)) ?

The integral of (2x)/((x+3)(3x+1)) is 1/12 (-ln|3x+1|+9ln|3x+9|)+C

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integral of (2x)/((x+3)(3x+1))

\int \frac{2 x}{( x + 3 ) ( 3 x + 1 )} d x

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

Solution steps

\int \frac{2 x}{( x + 3 ) ( 3 x + 1 )} d x

Take the constant out:

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

= 2 \cdot \int \frac{x}{( x + 3 ) ( 3 x + 1 )} d x

Apply u-substitution

= 2 \cdot \int \frac{u - 1}{3 u ( u + 8 )} d u

Take the constant out:

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

= 2 \cdot \frac{1}{3} \cdot \int \frac{u - 1}{u ( u + 8 )} d u

Expand

\frac{u - 1}{u ( u + 8 )} : \frac{u}{u ( u + 8 )} - \frac{1}{u ( u + 8 )}

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= 2 \cdot \frac{1}{3} (\int \frac{u}{u ( u + 8 )} d u - \int \frac{1}{u ( u + 8 )} d u)

\int \frac{u}{u ( u + 8 )} d u = ln ∣ u + 8 ∣

\int \frac{1}{u ( u + 8 )} d u = \frac{1}{8} ln ∣ u ∣ - \frac{1}{8} ln ∣ u + 8 ∣

= 2 \cdot \frac{1}{3} (ln ∣ u + 8 ∣ - (\frac{1}{8} ln ∣ u ∣ - \frac{1}{8} ln ∣ u + 8 ∣))

Substitute back

u = 3 x + 1

= 2 \cdot \frac{1}{3} (ln ∣ 3 x + 1 + 8 ∣ - (\frac{1}{8} ln ∣ 3 x + 1 ∣ - \frac{1}{8} ln ∣ 3 x + 1 + 8 ∣))

Simplify

2 \cdot \frac{1}{3} (ln ∣ 3 x + 1 + 8 ∣ - (\frac{1}{8} ln ∣ 3 x + 1 ∣ - \frac{1}{8} ln ∣ 3 x + 1 + 8 ∣)) : \frac{1}{12} (- ln ∣ 3 x + 1 ∣ + 9 ln ∣ 3 x + 9 ∣)

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

Add a constant to the solution

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

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

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

\frac{\partial}{\partial y} (x cos (x + 2 y))

f (x) = \frac{1}{x + 1}, at x = 4

\int \frac{1}{( x ^{2} - 4 )} d x

What is the integral of (2x)/((x+3)(3x+1)) ?
The integral of (2x)/((x+3)(3x+1)) is 1/12 (-ln|3x+1|+9ln|3x+9|)+C