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

The integral from-1 to 0 of 1/(x^2+2x+2) is pi/4

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

\int_{- 1}^{0} \frac{1}{x ^{2} + 2 x + 2} d x

\frac{π}{4}

Decimal

0.78539 \dots

Solution steps

\int_{- 1}^{0} \frac{1}{x ^{2} + 2 x + 2} d x

Complete the square

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

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

Apply u-substitution

: \int_{0}^{1} \frac{1}{u ^{2} + 1} d u

= \int_{0}^{1} \frac{1}{u ^{2} + 1} d u

Use the common integral:

\int \frac{1}{u ^{2} + 1} d u = arctan (u)

= [arctan (u)]_{0}^{1}

Compute the boundaries:

\frac{π}{4}

= \frac{π}{4}

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What is the integral from-1 to 0 of 1/(x^2+2x+2) ?
The integral from-1 to 0 of 1/(x^2+2x+2) is pi/4