What is #f(x) = int 1/((x+3)(x^2+4) dx# if #f(-1) = 0 #?

1 Answer
Apr 16, 2017

#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."#

Explanation:

#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.!