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

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

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

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

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

Solution steps

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

Apply u-substitution

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

Take the constant out:

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

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

Take the partial fraction of

\frac{u - 1}{u ( u + 1 )} : - \frac{1}{u} + \frac{2}{u + 1}

= \frac{1}{2} \cdot \int - \frac{1}{u} + \frac{2}{u + 1} 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

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

\int \frac{1}{u} d u = ln ∣ u ∣

\int \frac{2}{u + 1} d u = 2 ln ∣ u + 1 ∣

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

Substitute back

u = 2 x + 1

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

Simplify

\frac{1}{2} (- ln ∣ 2 x + 1 ∣ + 2 ln ∣ 2 x + 1 + 1 ∣) : \frac{1}{2} (- ln ∣ 2 x + 1 ∣ + 2 ln ∣ 2 x + 2 ∣)

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

Add a constant to the solution

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

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