What is the integral from-1 to 1 of (2x+|x|+1)^2 ?

The integral from-1 to 1 of (2x+|x|+1)^2 is 22/3

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integral from-1 to 1 of (2x+|x|+1)^2

\int_{- 1}^{1} (2 x + ∣ x ∣ + 1)^{2} d x

\frac{22}{3}

Decimal

7.33333 \dots

Solution steps

\int_{- 1}^{1} (2 x + ∣ x ∣ + 1)^{2} d x

Eliminate Absolutes

= \int_{- 1}^{0} (2 x - x + 1)^{2} d x + \int_{0}^{1} (2 x + x + 1)^{2} d x

\int_{- 1}^{0} (2 x - x + 1)^{2} d x = \frac{1}{3}

\int_{0}^{1} (2 x + x + 1)^{2} d x = 7

= \frac{1}{3} + 7

Simplify

\frac{1}{3} + 7 : \frac{22}{3}

= \frac{22}{3}

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What is the integral from-1 to 1 of (2x+|x|+1)^2 ?
The integral from-1 to 1 of (2x+|x|+1)^2 is 22/3