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If f(x)=4x2+4x3 then x+3f(x)dx is equal to 

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a
(1/4)f(x)+(2/3)log⁡|2x+1+f(x)|+C
b
(1/4)f(x)+(5/4)log⁡|2x+1+f(x)|+C
c
(1/3)f(x)+(2/3)log⁡|(x+3)+f(x)|+C
d
(2/3)f(x)+(1/3)log⁡|(x+3)+f(x)|

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detailed solution

Correct option is B

Put 2x+1=t, so that f(x)=t2−4 then I=14∫t+5t2−4dt=14t2−4+54log⁡t+t2−4+C=14f(x)+54log⁡|2x+1+f(x)|+C.


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