112(2x+3)6+k
13(2x+3)12+k
13(2x+3)32+k
112(2x+3)34+k
Correct answer is C
∫(2x+3)12δx
let u = 2x + 3, δyδx=2
δx=δu2
Now ∫(2x+3)12δx=∫u12.δx2
=12∫u12δu
=12u32×23+k
=13u32+k
=13(2x+3)32+k
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