How do you integrate #f(x)=(x^2+1)/((x^2+6)(x-4))# using partial fractions?

1 Answer
Mar 30, 2016

#F(x)=5/44ln(x^2+6)+10/(11sqrt6)arctan(x/sqrt6)+17/22ln|x-4|+c#

Explanation:

#f(x)=int(x^2+1)/((x^2+6)(x-4))dx#

Partial fraction decomposition is #(Ax+B)/(x^2+6)+C/(x-4)#

#(Ax+B)(x-4) + C(x^2+6) = x^2+0x+1#
From #x^2# terms: #A+C=1#
From #x^1# terms: #B-4A=0#
From #x^0# terms: #6C-4B=1#
Simultaneous solving yields #A=5/22#, #B=20/22# and #C=17/22#
Alternatively, use cover up rule where possible.

Thus the partial fraction decomposition is
#(5x+20)/(22(x^2+6))+17/(22(x-4))#

Thus #f(x)=int(5x+20)/(22(x^2+6))+17/(22(x-4))dx#

Algebraic manipulation yields:
#f(x)=5/44int(2x)/(x^2+6)dx+10/11int1/(x^2+6)dx+17/22int1/(x-4)dx#

Integrating yields:
#F(x)=5/44ln(x^2+6)+10/(11sqrt6)arctan(x/sqrt6)+17/22ln|x-4|+c#