Question

Partial Fractions\[\int\limits_{}^{}\frac{ dx }{ x^2(x^2-16) }\]

Answer 1

should I split up x^2-16

Answer 2

without partial fraction, there's a way, adjust the numerator,
1 = -16/-16
1 = (-1/16) (x^2-16 -x^2)
did you get this adjustment ?

Answer 3

no

Answer 4

i get the top part but not after that

Answer 5

we want numerator = x^2-16 -x^2 {=-16 } so that we can split the denominator
but the actual numerator =1
so i just multiplied and divided by -16

Answer 6

if you want to do partial fractions, then yes, you must split up (x^2 - 16) into (x+4)(x-4)

Answer 7

ok

Answer 8

but what hartnn is doing is pretty cool too.

\[\int\limits_{}^{} \frac{ dx }{ x^2(x^2 - 6) } \int\limits_{}^{}\left( -\frac{ 1 }{ 16 } \right)\frac{ x^2 - 16 - x^2 }{ x^2(x^2 - 16) } dx = -\frac{ 1 }{ 16 }\int\limits_{}^{}\frac{ 1 }{ x^2 } - \frac{ 1 }{ x^2 - 16 } dx\]

Answer 9

forgot the first equal sign lol

Answer 10

with partial fractions:
\[\int\limits_{}^{}\frac{ dx }{ x^2(x^2 - 16) } =>\frac{ 1 }{ x^2(x-4)(x+4) }= \frac{ A }{ x } + \frac{ B }{ x^2 } + \frac{ C }{ x-4 } + \frac{ D }{ x+4 }\]
so much longer than adjusting the numerator.

Answer 11

ok thanks you guys