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

The integral of 1/(x^2+4x+4) is -1/(x+2)+C

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

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

- \frac{1}{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 ^{2}} d u

= \int \frac{1}{u ^{2}} d u

Apply the Power Rule

: - \frac{1}{u}

= - \frac{1}{u}

Substitute back

u = x + 2

= - \frac{1}{x + 2}

Add a constant to the solution

= - \frac{1}{x + 2} + C

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