How do you integrate $int (2x+1)/((x-4)(x-1)(x+7))$ using partial fractions?

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Answer 1 · 2016-12-09T10:44:38

The answer is $=3/11ln(∣x-4∣)-1/8ln(∣x-1∣)-13/88ln(∣x+7∣)+C$

Let's do the decomposition into partial fractions

$(2x+1)/((x-4)(x-1)(x+7))=A/(x-4)+B/(x-1)+C/(x+7)$

$=(A(x-1)(x+7)+B(x-4)(x+7)+C(x-1)(x-4))/((x-4)(x-1)(x+7))$

Therefore,

$2x+1=A(x-1)(x+7)+B(x-4)(x+7)+C(x-1)(x-4)$

Let $x=1$ , $=>$ , $3=-24B$ , $=>$ , $B=-1/8$

Let $x=4$ , $=>$ , $9=33A$ , $=>$ , $A=3/11$

Let $x=-7$ , $=>$ , $-13=88C$ , $=>$ $C=-13/88$

So,

$(2x+1)/((x-4)(x-1)(x+7))=(3/11)/(x-4)-(1/8)/(x-1)-(13/88)/(x+7)$

$int((2x+1)dx)/((x-4)(x-1)(x+7))=3/11intdx/(x-4)-1/8intdx/(x-1)-13/88intdx/(x+7)$

$=3/11ln(∣x-4∣)-1/8ln(∣x-1∣)-13/88ln(∣x+7∣)+C$

How do you integrate int (2x+1)/((x-4)(x-1)(x+7)) int (2x+1)/((x-4)(x-1)(x+7)) using partial fractions?