What is the integral from-3 to-1 of 1/(x^2+6x+13) ?

The integral from-3 to-1 of 1/(x^2+6x+13) is pi/8

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integral from-3 to-1 of 1/(x^2+6x+13)

\int_{- 3}^{- 1} \frac{1}{x ^{2} + 6 x + 13} d x

\frac{π}{8}

Decimal

0.39269 \dots

Solution steps

\int_{- 3}^{- 1} \frac{1}{x ^{2} + 6 x + 13} d x

Complete the square

x^{2} + 6 x + 13 : (x + 3)^{2} + 4

= \int_{- 3}^{- 1} \frac{1}{( x + 3 ) ^{2} + 4} d x

Apply u-substitution

: \int_{0}^{2} \frac{1}{u ^{2} + 4} d u

= \int_{0}^{2} \frac{1}{u ^{2} + 4} d u

Apply Integral Substitution

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

Take the constant out:

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

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

Use the common integral:

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

= \frac{1}{2} [arctan (v)]_{0}^{1}

Compute the boundaries:

\frac{π}{4}

= \frac{1}{2} \cdot \frac{π}{4}

Simplify

= \frac{π}{8}

\int \frac{1}{( x + 3 ) x ^{2} + 6 x + 8} d x

\int (5 x^{4}) d x

7 (ln (y)) y^{^{'}} - x^{9} y = 0

(f (4 x))^{2} + 5

\frac{d}{d x} (lo g_{12} (x))

What is the integral from-3 to-1 of 1/(x^2+6x+13) ?
The integral from-3 to-1 of 1/(x^2+6x+13) is pi/8