What is the integral of |2x-1| ?

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

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integral of |2x-1|

\int ∣ 2 x - 1 ∣ d x

\frac{1}{4} ∣ 2 x - 1 ∣ (2 x - 1) + C

Solution steps

\int ∣ 2 x - 1 ∣ d x

Apply u-substitution

: \int ∣ u ∣ \frac{1}{2} d u

= \int ∣ u ∣ \frac{1}{2} 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 ∣ u ∣ d u

Use the common integral:

\int ∣ u ∣ d u = \frac{u ∣ u ∣}{2}

= \frac{1}{2} \cdot \frac{u ∣ u ∣}{2}

Substitute back

u = 2 x - 1

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

Simplify

\frac{1}{2} \cdot \frac{( 2 x - 1 ) ∣ 2 x - 1 ∣}{2} : \frac{1}{4} ∣ 2 x - 1 ∣ (2 x - 1)

= \frac{1}{4} ∣ 2 x - 1 ∣ (2 x - 1)

Add a constant to the solution

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

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\frac{d}{d x} (\frac{2 x}{4 + x})

\int \frac{1}{4 + t ^{2}} d t

\frac{\partial}{\partial x} (\frac{x - y}{x ^{2} + y ^{2} + 1})