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

The integral from 1 to 5 of 5/(x^2+4x+4) is 20/21

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

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

\frac{20}{21}

Decimal Notation

0.95238 \dots

Solution steps

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

Complete the square

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

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

Apply u-substitution

: \int_{3}^{7} \frac{1}{u ^{2}} d u

= 5 \cdot \int_{3}^{7} \frac{1}{u ^{2}} d u

Apply the Power Rule

: [- \frac{1}{u}]_{3}^{7}

= 5 [- \frac{1}{u}]_{3}^{7}

Compute the boundaries:

\frac{4}{21}

= 5 \cdot \frac{4}{21}

Simplify

= \frac{20}{21}

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

\frac{\partial}{\partial y} (\frac{x ^{2}}{1 + y ^{3}})

\frac{d}{d x} (a x e^{b x^{2}})

\int - \frac{arccot ( x )}{1 + x ^{2}} d x

\frac{\partial}{\partial x} (3 x^{2} + 6 x y)

What is the integral from 1 to 5 of 5/(x^2+4x+4) ?
The integral from 1 to 5 of 5/(x^2+4x+4) is 20/21