Question

Integrate by partial fraction
(x^2-5x+6)/((2x+3)(x^2+4))

Answer 1

let
(x^2-5x+6)/((2x+3)(x^2+4)) = A/(2x+3) + B/(x^2 + 4)
= (A(x^2 + 4) + B(2x+3) ) / (x^2-5x+6)/((2x+3)(x^2+4))

then x^2-5x+6 = (A(x^2 + 4) + B(2x+3) ) , which is an identity, true for all
defined values of x

let x = 0, ----> 6 = 4A + 3B
let x = 1 ----> 2 = 5A + 5B

multiply the first by 5 ---> 30 = 20A + 15B
multiply the 2nd by 3 ---> 6 = 15A + 15B
subtract them:
24 = 5A
A = 24/5
in 5A + 5B = 2
5(24/5) + 5B = 2
5B = -22
B = -22/5

so .....
(x^2-5x+6)/((2x+3)(x^2+4))
= 24/(5(2x+3)) - 22/(5(x^2 + 4)

So now to the integrals
the first part should be ok, just use natural logs to get
∫ (24/5) / (2x+3) dx = (12/5) ln (2x+3)

the second part will require some trig substitution.
Do you know how to do this ?

Answer 2

let
(x^2-5x+6)/((2x+3)(x^2+4)) = A/(2x+3) + (Bx+C)/(x^2 + 4)
following the steps above, this comes out to
A = 63/25
B = -19/25
C = -34/25
so
let
(x^2-5x+6)/((2x+3)(x^2+4)) = 1/25 (63/(2x+3) - (19x+34)/(x^2 + 4))
making the integral
1/50 (63ln(2x+3) - 19ln(x^2+4) - 34tan^{^-1}(x/2)) + C

Answer 3

Often we get more than one way to break apart a question
using partial fractions
just like:
34/75 = 23/150 + 3/10
34/75 = 2/15 + 8/25

Mine:
h ttps://www.wolframalpha.com/input?i=+24%2F%285%282x%2B3%29%29+-+22%2F%285%28x%5E2+%2B+4%29%29

ooblecks:
h ttps://www.wolframalpha.com/input?i=simplify+1%2F25+%2863%2F%282x%2B3%29+-+%2819x%2B34%29%2F%28x%5E2+%2B+4%29%29

both yielded the same rational expression, so both versions
are correct