8(2x+1)2+k
6(2x+1)2+k
18(2x+1)4+k
16(2x+1)4+k
Correct answer is C
Using substitution method, Let u=2x+1
dudx=2==>du=2dx==>dx=du2
=∫u32du=12∫u3du
=12(u44)=u48
∴
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