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Evaluate (2x+3)12δx...

Evaluate (2x+3)12δx

A.

112(2x+3)6+k

B.

13(2x+3)12+k

C.

13(2x+3)32+k

D.

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

=12u12δu

=12u32×23+k

=13u32+k

=13(2x+3)32+k