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#