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.!