What is the integral from 1 to 3 of 5/(x^2+4x+4) ?

The integral from 1 to 3 of 5/(x^2+4x+4) is 2/3

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integral from 1 to 3 of 5/(x^2+4x+4)

\int_{1}^{3} \frac{5}{x ^{2} + 4 x + 4} d x

\frac{2}{3}

Decimal

0.66666 \dots

Solution steps

\int_{1}^{3} \frac{5}{x ^{2} + 4 x + 4} d x

Take the constant out:

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

= 5 \cdot \int_{1}^{3} \frac{1}{x ^{2} + 4 x + 4} d x

Complete the square

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

= 5 \cdot \int_{1}^{3} \frac{1}{( x + 2 ) ^{2}} d x

Apply u-substitution

: \int_{3}^{5} \frac{1}{u ^{2}} d u

= 5 \cdot \int_{3}^{5} \frac{1}{u ^{2}} d u

Apply the Power Rule

: [- \frac{1}{u}]_{3}^{5}

= 5 [- \frac{1}{u}]_{3}^{5}

Compute the boundaries:

\frac{2}{15}

= 5 \cdot \frac{2}{15}

Simplify

= \frac{2}{3}

\frac{d}{d x} (3 x^{2} e^{x} + e^{x} (x^{3} + 8))

\frac{d}{d y} (\frac{y ^{3}}{3})

x \to 1 - lim (∣ x ∣)

y^{^{'}} = 0.0015 y (2700 - y)

\frac{d}{d x} ((\frac{x}{1 + x})^{5})

What is the integral from 1 to 3 of 5/(x^2+4x+4) ?
The integral from 1 to 3 of 5/(x^2+4x+4) is 2/3