How do you integrate #int (x+4)/(x^2 + 2x + 5)# using partial fractions?

We seek:

# I = int \ (x+4)/(x^2+2x+5) \ dx #

We not that the denominator does not factorise with real factors so use of Partial Fractions would not be appropriate. Instead we can decompose the integrand as follows:

# I = int \ (1/2(2x+2) +3)/(x^2+2x+5) \ dx #

Where we have manipulated the numerator in such a way that it is the derivative of the denominator, so that:

# I = int \ (1/2(2x+2))/(x^2+2x+5) +3/(x^2+2x+5) \ dx #
# \ \ = 1/2 \ int \ (2x+2)/(x^2+2x+5) \ dx +3 \ int 1/(x^2+2x+5) \ dx #

For the First Integral, #I_1#, say, we can use a substitution, Lte

# u = x^2+2x+5 => (du)/dx = 2x+2 #

Then substituting we get (omitting the constant of integration):

# I_1 = 1/2 \ int \ 1/u \ du #
# \ \ \ = 1/2 \ ln|u| #

And restoring the substitution, we get:

# I_1 = 1/2 \ ln |x^2+2x+5| #

Next, we consider the second integral, #I_2#, say, we complete the square of the denominator, thus:

# I_2 = 3 \ int 1/((x+1)^2-1^2+5) \ dx #
# \ \ \ = 3 \ int 1/((x+1)^2+4) \ dx #

And we can perform a substitution, Let:

# u = (x+1)/2 => (du)/dx = 1/2 #

And substituting we get (omitting the constant of integration):

# I_2 = 3 \ int 2/((2u)^2+4) \ du #
# \ \ \ = 3 \ int 2/(4u^2+4) \ du #
# \ \ \ = 3/2 \ int 1/(u^2+1) \ du #
# \ \ \ = 3/2 \ arctan u #

And restoring the substitution, we get:

# I_2 = 3/2 \ arctan((x+1)/2) #

Combining both result and incorporating the integration constant, we get:

# I = 1/2 \ ln |x^2+2x+5| + 3/2 \ arctan((x+1)/2) + C #