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

The integral of x/((2x+1)^2) is 1/4 (ln|2x+1|+1/(2x+1))+C

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

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

\frac{1}{4} (ln ∣ 2 x + 1 ∣ + \frac{1}{2 x + 1}) + C

Solution steps

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

Apply u-substitution

: \int \frac{u - 1}{4 u ^{2}} d u

= \int \frac{u - 1}{4 u ^{2}} d u

Take the constant out:

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

= \frac{1}{4} \cdot \int \frac{u - 1}{u ^{2}} d u

Expand

\frac{u - 1}{u ^{2}} : \frac{1}{u} - \frac{1}{u ^{2}}

= \frac{1}{4} \cdot \int \frac{1}{u} - \frac{1}{u ^{2}} 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}{4} (\int \frac{1}{u} d u - \int \frac{1}{u ^{2}} d u)

\int \frac{1}{u} d u = ln ∣ u ∣

\int \frac{1}{u ^{2}} d u = - \frac{1}{u}

= \frac{1}{4} (ln ∣ u ∣ - (- \frac{1}{u}))

Substitute back

u = 2 x + 1

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

Simplify

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

Add a constant to the solution

= \frac{1}{4} (ln ∣ 2 x + 1 ∣ + \frac{1}{2 x + 1}) + C

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What is the integral of x/((2x+1)^2) ?
The integral of x/((2x+1)^2) is 1/4 (ln|2x+1|+1/(2x+1))+C