What is the integral from-1 to 0 of 1/(4x^2+8x+8) ?

The integral from-1 to 0 of 1/(4x^2+8x+8) is pi/(16)

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integral from-1 to 0 of 1/(4x^2+8x+8)

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

\frac{π}{16}

Decimal

0.19634 \dots

Solution steps

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

\frac{1}{4 x ^{2} + 8 x + 8} = \frac{1}{4} \frac{1}{x ^{2} + 2 x + 2}

= \int_{- 1}^{0} \frac{1}{4} \cdot \frac{1}{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

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

Complete the square

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

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

Apply u-substitution

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

Use the common integral:

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

= \frac{1}{4} [arctan (u)]_{0}^{1}

Compute the boundaries:

\frac{π}{4}

= \frac{1}{4} \cdot \frac{π}{4}

Simplify

= \frac{π}{16}

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What is the integral from-1 to 0 of 1/(4x^2+8x+8) ?
The integral from-1 to 0 of 1/(4x^2+8x+8) is pi/(16)