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

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

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

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

\frac{2}{5}

Decimal Notation

0.4

Solution steps

\int_{1}^{3} \frac{3}{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

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

Complete the square

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

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

Apply u-substitution

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

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

Apply the Power Rule

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

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

Compute the boundaries:

\frac{2}{15}

= 3 \cdot \frac{2}{15}

Simplify

= \frac{2}{5}

sin (θ) + θ cos (θ)

x \to 0 lim (\frac{3 x ^{2} + 3}{3})

f (x) = sin (8 x)

x^{π} π^{x}

f (x) = 4 x^{\frac{3}{5}}

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