What is the integral of |x+1| ?

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

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integral of |x+1|

\int ∣ x + 1 ∣ d x

\frac{( x + 1 ) ∣ x + 1 ∣}{2} + C

Solution steps

\int ∣ x + 1 ∣ d x

Apply u-substitution

= \int ∣ u ∣ d u

Use the common integral:

\int ∣ u ∣ d u = \frac{u ∣ u ∣}{2}

= \frac{u ∣ u ∣}{2}

Substitute back

u = x + 1

= \frac{( x + 1 ) ∣ x + 1 ∣}{2}

Add a constant to the solution

= \frac{( x + 1 ) ∣ x + 1 ∣}{2} + C

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