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sammii2u Group TitleBest ResponseYou've already chosen the best response.0
\[\int\limits_{1}^{6} \frac{ x2 }{ x^25x14 }\]
 one year ago

slaaibak Group TitleBest ResponseYou've already chosen the best response.0
\[\int\limits_{1}^{6} {x2 \over (x+2)(x 7)} dx\] now do the whole partial fractions thing
 one year ago

sammii2u Group TitleBest ResponseYou've already chosen the best response.0
\[\frac{ A }{ x7 } + \frac{ B }{ x+2 }\] then getting everything with the same denominator I get.. \[x2 = A(x+2) + B (x7)\] then plugging in the different zeros, x=7 and x=2 I solve for A and B. \[ A = \frac{ 5 }{ 9 }\] \[ B = \frac{ 4}{ 9 }\] Now the overall equation looks like: \[= \frac{ 5 }{ 9 } \int\limits_{}^{} \frac{ 1 }{ x7 } dx + \frac{ 4 }{ 9 } \int\limits_{}^{} \frac{ 1 }{ x+2 } dx\] now what..? am I even doing this right?
 one year ago

Goten77 Group TitleBest ResponseYou've already chosen the best response.1
yes... but in this form now its put into the ln form....
 one year ago

Goten77 Group TitleBest ResponseYou've already chosen the best response.1
dw:1358132021880:dw
 one year ago

Goten77 Group TitleBest ResponseYou've already chosen the best response.1
you got it into the like ... i cant think of the name but its a form name.... but you did it right
 one year ago

sammii2u Group TitleBest ResponseYou've already chosen the best response.0
right. so it would be \[= \left[ \frac{ 5\ln x7 }{ 9 } + \frac{ 4\ln x+2 }{ 9 } \right] _{1} ^{6}\] and then I evaluate it at x=6 and then subtract x=1 ?
 one year ago

slaaibak Group TitleBest ResponseYou've already chosen the best response.0
yeah. nice job.
 one year ago
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