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\[\int\limits_{}^{}(27x^2)\div (3x+4)^3 dx\]

oh ok

Do we have to do long division? Can we use partial fractions instead?

It actually says do long division?

Oh that is true

good eye

ok, so how would you do it with partial fractions?

Would you like a step by step?

if you could, that would help a lot!

No problem at all, first: how familiar are you with partial fractions?

i'm not familiar with the name but once i see it i will probably remember..

is there any other way to solve it?

ok now it looks familiar!!

so how would i get A B and C?

Is it ok if i wrote it:\[27\int\limits_{}^{}1\div(9(3x+4)-8\div9(3x+4)^2+16\div9(3x+4)^3 dx\]

i took out constant first
are my numbers correct?

well what would i use for my u value if i were to use u-substitution?

so that i only have to substitute once.

For all of them let u=3x+4; (1/3)du=dx

but i will still have to substitute again after

i got \[(u^2\div2)+(8\div27u)-(8\div27u^2)+ c] is that correct?

\[(u^2/2)+(8/27u)-(8/27u^2)+c\]**

then plug in u-value?