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

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

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

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

\frac{2}{3}

Decimal Notation

0.66666 \dots

Solution steps

\int_{1}^{4} \frac{4}{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

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

Complete the square

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

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

Apply u-substitution

: \int_{3}^{6} \frac{1}{u ^{2}} d u

= 4 \cdot \int_{3}^{6} \frac{1}{u ^{2}} d u

Apply the Power Rule

: [- \frac{1}{u}]_{3}^{6}

= 4 [- \frac{1}{u}]_{3}^{6}

Compute the boundaries:

\frac{1}{6}

= 4 \cdot \frac{1}{6}

Simplify

= \frac{2}{3}

x \to 0 lim (- 2)

\frac{d}{d x} (8 cos (3 x))

\int x I n (x) d x

\frac{d}{d x} (\frac{2000 x}{4 x + 75})

f (x) = 6 x arcsin (x)

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