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

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

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

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

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

Solution steps

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

Apply u-substitution

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

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

Apply u-substitution

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

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

Use the common integral:

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

= arctan (v)

Substitute back

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

Expand

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

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

Add a constant to the solution

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

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