What is the integral of (x+1)/(x^2+4x+6) ?

The integral of (x+1)/(x^2+4x+6) is 1/2 ln|x^2+4x+6|-1/(sqrt(2))arctan(1/(sqrt(2))(x+2))+C

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integral of (x+1)/(x^2+4x+6)

\int \frac{x + 1}{x ^{2} + 4 x + 6} d x

\frac{1}{2} ln x^{2} + 4 x + 6 - \frac{1}{2} arctan (\frac{1}{2} (x + 2)) + C

Solution steps

\int \frac{x + 1}{x ^{2} + 4 x + 6} d x

Complete the square

x^{2} + 4 x + 6 : (x + 2)^{2} + 2

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

Apply u-substitution

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

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

Expand

\frac{u - 1}{u ^{2} + 2} : \frac{u}{u ^{2} + 2} - \frac{1}{u ^{2} + 2}

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

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

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

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

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

Substitute back

u = x + 2

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

Simplify

\frac{1}{2} ln (x + 2)^{2} + 2 - \frac{1}{2} arctan (\frac{x + 2}{2}) : \frac{1}{2} ln x^{2} + 4 x + 6 - \frac{1}{2} arctan (\frac{1}{2} (x + 2))

= \frac{1}{2} ln x^{2} + 4 x + 6 - \frac{1}{2} arctan (\frac{1}{2} (x + 2))

Add a constant to the solution

= \frac{1}{2} ln x^{2} + 4 x + 6 - \frac{1}{2} arctan (\frac{1}{2} (x + 2)) + C

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What is the integral of (x+1)/(x^2+4x+6) ?
The integral of (x+1)/(x^2+4x+6) is 1/2 ln|x^2+4x+6|-1/(sqrt(2))arctan(1/(sqrt(2))(x+2))+C