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

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

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

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

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

Solution steps

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

Complete the square

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

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

Apply u-substitution

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

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

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

Apply u-substitution

: \int \frac{1}{v ^{2} + 1} d v

= \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} arctan (v)

Substitute back

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

Simplify

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

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

Add a constant to the solution

= \frac{1}{2} arctan (x^{2} + 4 x + 3) + C

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