How do you use partial fractions to find the integral #int (x^2+x+3)/(x^4+6x^2+9)dx#?

2 Answers

See the answer below:
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Jan 18, 2017

#1/sqrt(3)tan^(-1)(x/sqrt(3))-1/(2(x^2+3))+C#

Explanation:

Noticing that the denominator is a perfect square, we see there is no need to use partial fractions.

#(x^2+x+3)/(x^4+6x^2+9)=(x^2+3+x)/(x^2+3)^2=cancel(x^2+3)/(x^2+3)^cancel(2)+x/(x^2+3)^2#
#=1/(x^2+3)+x/(x^2+3)^2#
The first term is a standard integral of the form #1/(x^2+a^2)# with #a=sqrt(3)#, yielding #(1/sqrt(3))tan^(-1)(x/sqrt(3))#.

The second term is of the form where the the numerator is the derivative of the contents of the brackets in the denominator, save for a factor of #2#, and so can be integrated at sight. (Alternatively substitute #u=x^2+3#, #dx/(du)=1/(2x)# to give #1/2int u^(-2)du#.)

In general, #d/(dx)(1/(f(x)))=-(f prime(x))/(f(x))^2# so #int (f prime(x))/(f(x))^2dx=-1/(f(x))#. In this case, #f(x)=x^2+3# and #f prime(x)=2x#.