How do you integrate $int(x+1)/((6x^2+4)(x-5))$ using partial fractions?

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Answer 1 · 2017-10-02T17:41:00

See below.

1) I used basic fractions method.

Inıtıially, I decomposed integrand into basic fractions,

$(x+1)/[(6x^2+4)*(x-5)]=A/(x-5)+(Bx+C)/(6x^2+4)$

After expanding denominator,

$A*(6x^2+4)+(Bx+C)*(x-5)=x+1$

$(6A+B)*x^2+(C-5B)*x+4A-5C=x+1$

After equating coefficients, I found

$6A+B=0$ , $C-5B=1$ and $4A-5C=1$ equations.

After solving them, $A=3/77, B=-18/77 and C=-13/77$

Thus,

$int ((x+1)dx)/[(6x^2+4)*(x-5)]$

= $3/77$ $int dx/(x-5)$ - $1/77$ $int ((18x+13)*dx)/(6x^2+4)$

= $3/77*ln(x-5)$ - $3/154$ $int (12x*dx)/(6x^2+4)$ - $13/77$ $int dx/(6x^2+4)$

= $3/77*ln(x-5)-3/154*ln(6x^2+4)-(13sqrt(6))/924*arctan((3x)/sqrt(6))+C$

How do you integrate int(x+1)/((6x^2+4)(x-5))int(x+1)/((6x^2+4)(x-5)) using partial fractions?