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

The integral of 1/(4x^2+4x+4) is 1/(2sqrt(3))arctan((2x+1)/(sqrt(3)))+C

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

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

\frac{1}{2 3} arctan (\frac{2 x + 1}{3}) + C

Solution steps

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

Complete the square

4 x^{2} + 4 x + 4 : 4 (x + \frac{1}{2})^{2} + 3

= \int \frac{1}{4 ( x + \frac{1}{2} ) ^{2} + 3} d x

Apply u-substitution

: \int \frac{1}{4 u ^{2} + 3} d u

= \int \frac{1}{4 u ^{2} + 3} d u

Apply Integral Substitution

= \int \frac{1}{2 3 ( v ^{2} + 1 )} d v

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= \frac{1}{2 3} \cdot \int \frac{1}{v ^{2} + 1} d v

Use the common integral:

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

= \frac{1}{2 3} arctan (v)

Substitute back

= \frac{1}{2 3} arctan (\frac{2}{3} (x + \frac{1}{2}))

Simplify

\frac{1}{2 3} arctan (\frac{2}{3} (x + \frac{1}{2})) : \frac{1}{2 3} arctan (\frac{2 x + 1}{3})

= \frac{1}{2 3} arctan (\frac{2 x + 1}{3})

Add a constant to the solution

= \frac{1}{2 3} arctan (\frac{2 x + 1}{3}) + C

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What is the integral of 1/(4x^2+4x+4) ?
The integral of 1/(4x^2+4x+4) is 1/(2sqrt(3))arctan((2x+1)/(sqrt(3)))+C