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

1 Answer
benjidunn
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

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