What is the integral from 1 to 6 of 4/(x^2+4x+4) ?

The integral from 1 to 6 of 4/(x^2+4x+4) is 5/6

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integral from 1 to 6 of 4/(x^2+4x+4)

\int_{1}^{6} \frac{4}{x ^{2} + 4 x + 4} d x

\frac{5}{6}

Decimal Notation

0.83333 \dots

Solution steps

\int_{1}^{6} \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}^{6} \frac{1}{x ^{2} + 4 x + 4} d x

Complete the square

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

= 4 \cdot \int_{1}^{6} \frac{1}{( x + 2 ) ^{2}} d x

Apply u-substitution

: \int_{3}^{8} \frac{1}{u ^{2}} d u

= 4 \cdot \int_{3}^{8} \frac{1}{u ^{2}} d u

Apply the Power Rule

: [- \frac{1}{u}]_{3}^{8}

= 4 [- \frac{1}{u}]_{3}^{8}

Compute the boundaries:

\frac{5}{24}

= 4 \cdot \frac{5}{24}

Simplify

= \frac{5}{6}

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

\frac{\partial}{\partial x} (\frac{( 3 x ^{4} - 2 x y )}{y ^{2}})

\int \frac{2 x}{x ^{4} + 4} d x

\frac{d y}{d x} = \frac{6 x}{y}

x^{2} + 1

What is the integral from 1 to 6 of 4/(x^2+4x+4) ?
The integral from 1 to 6 of 4/(x^2+4x+4) is 5/6