`"If"int(dx)/(x^(3)(1+x^(6))^(2/3))=xf(x)(1+x^(6))^(1/3)+C` where, C is a constant of integration, then the function f(x) is equal toA. `-(1)/(6x^(3))`B. `-(1)/(2x^(3))`C. `-(1)/(2x^(2))`D. `(3)/(x^(2))`

`"If"int(dx)/(x^(3)(1+x^(6))^(2/3))=xf(x)(1+x^(6))^(1/3)+C` where, C i

1.	`"If"int(dx)/(x^(3)(1+x^(6))^(2/3))=xf(x)(1+x^(6))^(1/3)+C` where, C is a constant of integration, then the function f(x) is equal toA. `-(1)/(6x^(3))`B. `-(1)/(2x^(3))`C. `-(1)/(2x^(2))`D. `(3)/(x^(2))`
Answer» Correct Answer - B Let `I = int(dx)/(x^(3)(1+x^(6))^(2//3))` `=int(dx)/(x^(3)x^(4)((1)/(x^(6))+1)^(2//3))=int(dx)/(x^(7)((1)/(x^(6))+1)^(2//3))` Now, put `(1)/(x^(6))+1 = t^(3)` `rArr" " -(6)/(x^(7))dx = 3t^(2)dt` `rArr" "(dx)/(x^(7))=-(t^(2))/(2)dt` So, `I=int(-(1)/(2)t^(2)dt)/(t^(2))= -(1)/(2)int dt` `= -(1)/(2)t+C = -(1)/(2)((1)/(x^(6))+1)^(1//3)+C" "[therefore t^(3)=(1)/(x^(6))+1]` `= -(1)/(2)(1)/(x^(2))(1+x^(6))^(1//3)+C` `= xf(x)(1+x^(6))^(1//3)+C" "["given"]` On comparing both sides, we get `f(x) = -(1)/(2x^(3))`

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