How do you integrate $int (x+5)/((x+3)(x-2)(x-7))$ using partial fractions?

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Answer 1 · 2017-10-05T16:42:33

The answer is $=1/25ln(|x+3|)-7/25ln(|x-2|)+6/25ln(|x-7|)+C$

Perform the decomposition into partial fractions

$(x+5)/((x+3)(x-2)(x-7))=A/(x+3)+B/(x-2)+C/(x-7)$

$=(A(x-2)(x-7)+B(x+3)(x-7)+C(x+3)(x-2))/((x+3)(x-2)(x-7))$

The denominators are the same, compare the numerators

$(x+5)=(A(x-2)(x-7)+B(x+3)(x-7)+C(x+3)(x-2))$

Let $x=-3$ , $=>$ , $2=50A$ , $A=1/25$

Let $x=2$ , $=>$ , $7=-25B$ , $B=-7/25$

Let $x=7$ , $=>$ , $12=50C$ , $C=6/25$

Therefore,

$(x+5)/((x+3)(x-2)(x-7))=(1/25)/(x+3)+(-7/25)/(x-2)+(6/25)/(x-7)$

So, the integral is

$int((x+5)dx)/((x+3)(x-2)(x-7))=int(1/25dx)/(x+3)+int(-7/25dx)/(x-2)+int(6/25dx)/(x-7)$

$=1/25ln(|x+3|)-7/25ln(|x-2|)+6/25ln(|x-7|)+C$

How do you integrate int (x+5)/((x+3)(x-2)(x-7)) int (x+5)/((x+3)(x-2)(x-7)) using partial fractions?