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find the integral (4x^3+2/2x^4+4x+3) dx
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\[\int\limits_{}^{}\frac{4x ^{3}+2}{2x ^{4}+4x+3}dx\]
I'm thinking a u-sub of u=2x^{4}+4x+3 will work, right?
my answer choices are \[2 \ln \left| 2x ^{4}+4x+3 \right|+C\] , \[\frac{-2}{(2x ^{4}+4x+3)^{2}}+C\] , \[1/2\ln \left| 2x ^{4}+4x+3 \right|+C\] , \[\frac{-1}{2(2x ^{4}+4x+3)^{2}}+C\]
we gotta try partial fraction
okay where do we go from there
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It's actually a simple U-sub
its 3rd choice do you need to show work?
let u=2x^4 +4x + 3 evalute du, du = 8x^3 + 4 = 1/2 (4x^3 +2) then the rest is a general natural log integral
are you sure it is the third choice? no i just need the correct answer
yep i used u substitution just like wattagwan
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sweet thank you so much
no problem
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