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

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

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integral from 1 to 2 of 3/(x^2+2x+2)

\int_{1}^{2} \frac{3}{x ^{2} + 2 x + 2} d x

3 (arctan (3) - arctan (2))

Decimal

0.42569 \dots

Solution steps

\int_{1}^{2} \frac{3}{x ^{2} + 2 x + 2} d x

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= 3 \cdot \int_{1}^{2} \frac{1}{x ^{2} + 2 x + 2} d x

Complete the square

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

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

Apply u-substitution

= 3 \cdot \int_{2}^{3} \frac{1}{u ^{2} + 1} d u

Use the common integral:

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

= 3 [arctan (u)]_{2}^{3}

Compute the boundaries:

arctan (3) - arctan (2)

= 3 (arctan (3) - arctan (2))

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