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

The integral of 1/(sqrt(x^2+4x+4)) is ln|x+2|+C

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integral of 1/(sqrt(x^2+4x+4))

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

ln ∣ x + 2 ∣ + C

Solution steps

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

Complete the square

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

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

Apply u-substitution

= \int \frac{1}{u} d u

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= ln ∣ u ∣

Substitute back

u = x + 2

= ln ∣ x + 2 ∣

Add a constant to the solution

= ln ∣ x + 2 ∣ + C

\frac{\partial}{\partial y} (\frac{5 y}{x})

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

\frac{d}{d x} ((4 x^{3} + 1)^{3})

a re a 5 - x^{2}, sin (x), - 2.384677, 2.0252116

\int 3 x^{3} + 1 x^{5} d x

What is the integral of 1/(sqrt(x^2+4x+4)) ?
The integral of 1/(sqrt(x^2+4x+4)) is ln|x+2|+C