FInd the indefinite integral using substitution. let u = 2x^2 + 3x + 9. (4x + 3)/ (2x^2 + 3x + 9)^(7/3)dx. can someone please give me the steps and answer . im in desperate need

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FInd the indefinite integral using substitution. let u = 2x^2 + 3x + 9. (4x + 3)/ (2x^2 + 3x + 9)^(7/3)dx. can someone please give me the steps and answer . im in desperate need

Mathematics
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du = (7/3)*(4x+3)^(4/3)
Do you see that?
Actually I take that back. Try [(7/3)(2x^2+3x+9)^(4/3)] * (4x+3) dx

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Other answers:

= du
I got (-3/4)[(2x^2)+3x+9)^(-3/4)]+C
Oh wait sorry. It's gettin close to bed time lol. Just ignore me. good grief
you just have to take the derivative of 2x^2+3x+9 and you get du=4x+3 dx...then you can substitute for the numberator of the integrand...then you can take the integral and back substitute u=2x^2+3x+9 after that.
numerator*
\[\int\limits_{?}^{?}( u^(-7/3) du\]
you can then take the integral of that
and then back substitute u to get the answer

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