What is the integral of s/(s^2+6s+11) ?

The integral of s/(s^2+6s+11) is 1/2 ln|s^2+6s+11|-3/(sqrt(2))arctan(1/(sqrt(2))(s+3))+C

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integral of s/(s^2+6s+11)

\int \frac{s}{s ^{2} + 6 s + 11} d s

\frac{1}{2} ln s^{2} + 6 s + 11 - \frac{3}{2} arctan (\frac{1}{2} (s + 3)) + C

Solution steps

\int \frac{s}{s ^{2} + 6 s + 11} d s

Complete the square

s^{2} + 6 s + 11 : (s + 3)^{2} + 2

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

Apply u-substitution

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

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

Expand

\frac{u - 3}{u ^{2} + 2} : \frac{u}{u ^{2} + 2} - \frac{3}{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{3}{u ^{2} + 2} d u

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

\int \frac{3}{u ^{2} + 2} d u = \frac{3}{2} arctan (\frac{u}{2})

= \frac{1}{2} ln u^{2} + 2 - \frac{3}{2} arctan (\frac{u}{2})

Substitute back

u = s + 3

= \frac{1}{2} ln (s + 3)^{2} + 2 - \frac{3}{2} arctan (\frac{s + 3}{2})

Simplify

\frac{1}{2} ln (s + 3)^{2} + 2 - \frac{3}{2} arctan (\frac{s + 3}{2}) : \frac{1}{2} ln s^{2} + 6 s + 11 - \frac{3}{2} arctan (\frac{1}{2} (s + 3))

= \frac{1}{2} ln s^{2} + 6 s + 11 - \frac{3}{2} arctan (\frac{1}{2} (s + 3))

Add a constant to the solution

= \frac{1}{2} ln s^{2} + 6 s + 11 - \frac{3}{2} arctan (\frac{1}{2} (s + 3)) + C

f (x) = 4 x - 3

\int - 2 cos (2 x) d x

x \to \frac{4}{π} lim (cos (x))

\frac{d}{d x} (\frac{1}{8} x^{8} - x^{4})

What is the integral of s/(s^2+6s+11) ?
The integral of s/(s^2+6s+11) is 1/2 ln|s^2+6s+11|-3/(sqrt(2))arctan(1/(sqrt(2))(s+3))+C