What is the integral from-2 to 0 of 2/(x^2+4x+8) ?

The integral from-2 to 0 of 2/(x^2+4x+8) is pi/4

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integral from-2 to 0 of 2/(x^2+4x+8)

\int_{- 2}^{0} \frac{2}{x ^{2} + 4 x + 8} d x

\frac{π}{4}

Decimal

0.78539 \dots

Solution steps

\int_{- 2}^{0} \frac{2}{x ^{2} + 4 x + 8} d x

Take the constant out:

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

= 2 \cdot \int_{- 2}^{0} \frac{1}{x ^{2} + 4 x + 8} d x

Complete the square

x^{2} + 4 x + 8 : (x + 2)^{2} + 4

= 2 \cdot \int_{- 2}^{0} \frac{1}{( x + 2 ) ^{2} + 4} d x

Apply u-substitution

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

Apply Integral Substitution

= 2 \cdot \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

= 2 \cdot \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)

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

Simplify

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

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

Compute the boundaries:

\frac{π}{4}

= \frac{π}{4}

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What is the integral from-2 to 0 of 2/(x^2+4x+8) ?
The integral from-2 to 0 of 2/(x^2+4x+8) is pi/4