How do you integrate $int (2x+1)/((x+8)(x-2)(x-3))$ using partial fractions?

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Answer 1 · 2017-01-31T16:13:55

The answer is $=-3/22ln(|x+8|)-1/2ln(|x-2|)+7/11ln(|x-3|)+C$

Let's perform the decomposition into partial fractions

$(2x+1)/((x+8)(x-2)(x-3))=A/(x+8)+B/(x-2)+C/(x-3)$

$=(A(x-2)(x-3)+B(x+8)(x-3)+C(x+8)(x-2))/((x+8)(x-2)(x-3))$

As the denominators are the same, we can equalise the numerators

$(2x+1)=A(x-2)(x-3)+B(x+8)(x-3)+C(x+8)(x-2)$

Let $x=-8$ , $=>$ , $-15=110A$ , $=>$ , $A=-3/22$

Let $x=2$ , $=>$ , $5=-10B$ , $=>$ , $B=-1/2$

Let $x=3$ , $=>$ , $7=11C$ , $=>$ , $C=7/11$

Therefore,

$(2x+1)/((x+8)(x-2)(x-3))=(-3/22)/(x+8)+(-1/2)/(x-2)+(7/11)/(x-3)$

So,

$int((2x+1)dx)/((x+8)(x-2)(x-3))=-3/22intdx/(x+8)-1/2intdx/(x-2)+7/11intdx/(x-3)$

$=-3/22ln(|x+8|)-1/2ln(|x-2|)+7/11ln(|x-3|)+C$

How do you integrate int (2x+1)/((x+8)(x-2)(x-3)) int (2x+1)/((x+8)(x-2)(x-3)) using partial fractions?