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

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

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integral of 1/(4x^2+4x+3)

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

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

Solution steps

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

Complete the square

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

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

Apply u-substitution

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

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

Apply Integral Substitution

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

Take the constant out:

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

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

\frac{1}{2 v ^{2} + 2} = \frac{1}{2} \frac{1}{v ^{2} + 1}

= \frac{1}{2} \cdot \int \frac{1}{2} \cdot \frac{1}{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} \cdot \frac{1}{2} \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} \cdot \frac{1}{2} arctan (v)

Substitute back

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

Simplify

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

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

Add a constant to the solution

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

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