What is the integral from-2 to 0 of 5/(x^2+4x+8) ?

The integral from-2 to 0 of 5/(x^2+4x+8) is (5pi)/8

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral from-2 to 0 of 5/(x^2+4x+8)

\int_{- 2}^{0} \frac{5}{x ^{2} + 4 x + 8} d x

\frac{5 π}{8}

Decimal

1.96349 \dots

Solution steps

\int_{- 2}^{0} \frac{5}{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

= 5 \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

= 5 \cdot \int_{- 2}^{0} \frac{1}{( x + 2 ) ^{2} + 4} d x

Apply u-substitution

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

= 5 \cdot \int_{0}^{2} \frac{1}{u ^{2} + 4} d u

Apply Integral Substitution

= 5 \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

= 5 \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)

= 5 \cdot \frac{1}{2} [arctan (v)]_{0}^{1}

Simplify

5 \cdot \frac{1}{2} [arctan (v)]_{0}^{1} : \frac{5}{2} [arctan (v)]_{0}^{1}

= \frac{5}{2} [arctan (v)]_{0}^{1}

Compute the boundaries:

\frac{π}{4}

= \frac{5}{2} \cdot \frac{π}{4}

Simplify

= \frac{5 π}{8}

x \to 4 lim (- x^{2} + 3 x - 7)

\frac{d}{d x} (\frac{π}{5})

\int x^{2} + 2 x - 15 d x

x^{2} - 43, - 2 x + 5, - 8, 6

x^{^{'''}} = 0

What is the integral from-2 to 0 of 5/(x^2+4x+8) ?
The integral from-2 to 0 of 5/(x^2+4x+8) is (5pi)/8