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Mathematics 15 Online
OpenStudy (anonymous):

Help? Step-by-step? integral from 0 to infinity of (1)/(x^2+7x+12)dx

OpenStudy (yrelhan4):

x^2+7x+12 what can you write this as?

OpenStudy (anonymous):

(x+4)(x+3)

OpenStudy (yrelhan4):

now can you use integration by parts?

OpenStudy (anonymous):

@yrelhan4 @monroe17 no need for integration by parts can we write the term in numerator as (x+4) - (x+3)??????

OpenStudy (yrelhan4):

yup.

OpenStudy (anonymous):

^huh?

OpenStudy (yrelhan4):

explain your method to her nik. its much easier. good job.

OpenStudy (anonymous):

cool integral of 1 /(x+4)(x+3) = integral of (x+4) - (x+3)/ (x+4)(x+3) did u understand this step? @monroe17

OpenStudy (anonymous):

is the numerator minus(x+3) because it was originally in the denominator and your moving it to the numerator? and the denominator stays the same?

OpenStudy (anonymous):

no that's not right ;/

OpenStudy (anonymous):

why is it not right?

OpenStudy (anonymous):

because if that was so, wouldn't all of the numerator be negative? not just -(x+3)?

OpenStudy (anonymous):

omg it is { (x+4) - (x+3) }/ (x+4)(x+3)

OpenStudy (anonymous):

? I'm confused as to why.

OpenStudy (anonymous):

in order to make integration simpler , it is one of the techniques because the second step is as follow integral of x+4/(x+4)(x+3) - x+3/(x+4)(x+3) then u can cut the like terms and see integration become very simple

OpenStudy (anonymous):

@monroe17 can u write 1 = 4 - 3 ?

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