How do you find $int (12x+34)/(x^2+6x+8)$ ?

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Answer 1 · 2015-11-07T21:11:09

$7ln|x+4|+5ln|x+2|+C$

I will use the method of partial fractions to solve this integral.

Since the denominator is linear factors, we may write it as

$(12x+34)/(x^2+6x+8)=A/(x+4)+B/(x+2)=(A(x+2)+B(x+4))/((x+4)(x+2))$

$thereforeAx+2A+Bx+4B=12x+34$

$therefore(A+B)x+(2A+4B)=12x+34$

$therefore$ , by comparing terms, $A+B=12and2A+4B=34$

Solving this system of linear equations yields $A=7and B=5$ .

$thereforeint(12x+34)/(x^2+6x+8)=int7/(x+4)dx=int5/(x+2)dx$

$=7ln|x+4|+5ln|x+2|+C$

How do you find int (12x+34)/(x^2+6x+8)int (12x+34)/(x^2+6x+8)?