What is the integral from 0 to 1 of (1+2x+x^2)^2 ?

The integral from 0 to 1 of (1+2x+x^2)^2 is 31/5

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integral from 0 to 1 of (1+2x+x^2)^2

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

\frac{31}{5}

Decimal

6.2

Solution steps

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

Complete the square

1 + 2 x + x^{2} : (x + 1)^{2}

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

Apply u-substitution

: \int_{1}^{2} u^{4} d u

= \int_{1}^{2} u^{4} d u

Apply the Power Rule

: [\frac{u ^{5}}{5}]_{1}^{2}

= [\frac{u ^{5}}{5}]_{1}^{2}

Compute the boundaries:

\frac{31}{5}

= \frac{31}{5}

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What is the integral from 0 to 1 of (1+2x+x^2)^2 ?
The integral from 0 to 1 of (1+2x+x^2)^2 is 31/5