Find the antiderivative of (6x)/(x^2 - x - 2)
you must fisrt factor the denominator and use partial fractions. \[= \int\limits_{}^{}\frac{ 6x }{ (x-2)(x+1) } dx = \int\limits_{}^{} [\frac{ A }{ x-2 } + \frac{ B }{ x+1 }] dx\] let me know if i should keep going
okay
okay that's enough or okay i should continue?
continue
partial fractions says that: \[\frac{ A }{ x-2 } + \frac{ B }{ x+1 } = \frac{ 6x }{ (x-2)(x+1) }\] this should come as no shock if you have ever added fractions on paper. you have to make the denominators equal to add them. this is what we will do. we must cross multiply to make the denominators the same: \[\frac{ A }{ x-2 }\frac{ x+1 }{ x+1 } + \frac{ B }{ x+1 }\frac{ x-2 }{ x-2 } = \frac{ 6x }{ (x-2)(x+1) }\] \[ =\frac{ Ax + A + Bx - 2B }{ (x-2)(x+1) } = \frac{ 6x }{ (x-2)(x+1) }\] we can therefore deduce, by comparing numerators of the equivlalent equations, that: A + B = 6 and A -2B = 0 solving for A and B: A = 4 and B = 2 you are now dealing with \[\int\limits_{}^{} \frac{ 4 }{ (x-2) } + \frac{ 2 }{ (x+1) } dx\]
okay
which is 4 ln| x-2| + 2 ln | x+1| + C
okay thanks
glad i could help. i hope you learned from this, you will be using this later :)
yes i have :)
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