#f(x)=int1/{(x+3)(x^2+4)}dx.#
We decompose #1/{(x+3)(x^2+4)}# using the Method Of Parial
Decomposition :
Suppose that, for some, #A,B,C in RR,# we have,
# 1/{(x+3)(x^2+4)}=A/(x+3)+(Bx+C)/(x^2+4)....(ast).#
#:. 1/{(x+3)(x^2+4)}={A(x^2+4)+(Bx+C)(x+3)}/{(x+3)(x^2+4)}.#
Using Heavisides Method, we get, #A=[1/(x^2+4)]_(x+3=0).#
#:. A=1/{(-3)^2+4}=1/13.#
#:., (ast) rArr 1/{(x+3)(x^2+4)}-(1/13)/(x+3)=(Bx+C)/(x^2+4), i.e., #
#(Bx+C)/(x^2+4)={1-1/13(x^2+4)}/{(x+3)(x^2+4)}=(1/13(9-x^2))/{(x+3)(x^2+4)}, or,#
#(Bx+C)/(x^2+4)=(1/13(3-x))/(x^2+4) rArr C=3/13, B=-1/13.#
#"Thus, "1/{(x+3)(x^2+4)}=(1/13)/(x+3)+{(-1/13)x+3/13}/(x^2+4).#
#"Thereore, "int 1/{(x+3)(x^2+4)}dx=1/13int1/(x+3)dx#
#+int{(-1/13)x+3/13}/(x^2+4)dx#
#=1/13ln|x+3|-1/13*1/2int(2x)/(x^2+4)dx+3/13int1/(x^2+2^2)dx,#
#=1/13ln|x+3|-1/26int{d/dx(x^2+4)}/(x^2+4)dx+3/13*1/2arc tan(x/2),#
#=1/13ln|x+3|-1/26ln(x^2+4)+3/26arc tan(x/2)+c.#
#:.f(x)=1/13ln|x+3|-1/26ln(x^2+4)+3/26arc tan(x/2)+c#
Finally, to determine #c,# we use the Initial Condition : f(-1)=0.#
#f(-1)=0 rArr 1/13ln|-1+3|-1/26ln((-1)^2+4)+3/26arc tan(-1/2)+c=0.#
#:. 1/13ln2-1/26ln5-3/26 arc tan(1/2)+c=0.#
#:. c=1/26ln5+3/26arc tan(1/2)-1/13ln2.#
#"Finally, therefore, "f(x)=1/13ln|x+3|-1/26ln(x^2+4)+3/26arc tan(x/2)+1/26ln5+3/26arc tan(1/2)-1/13ln2."#
We can say, #1/26ln5-1/13ln2=1/26(ln5-2ln2)=1/26ln(5/4).#
Enjoy Maths.!
