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`int(3x^(13)+2x^(11))/((2x^4+3x^2+1)^4)dx`A. `(x^(4))/(6(2x^(4)+3x^(2)+1)^(3))+C`B. `(x^(12))/(6(2x^(4)+3x^(2)+1)^(3))+C`C. `(x^(4))/((2x^(4)+3x^(2)+1)^(3))+C`D. `(x^(12))/((2x^(4)+3x^(2)+1)^(3))+C` |
Answer» Correct Answer - B Let `I=int(3x^(13)+2x^(11))/((2x^(4)+3x^(2)+1)^(4))dx = int((3)/(x^(3))+(2)/(x^(5)))/((2+(3)/(x^(2))+(1)/(x^(4)))^(4))dx` [on dividing numerator and denominator by `x^(16)`] Now, put `2 + (3)/(x^(2))+(1)/(x^(4))=t` `rArr ((-6)/(x^(3))-(4)/(x^(5)))dx = dt rArr ((3)/(x^(3))+(2)/(x^(5)))dx = - (dt)/(2)` So, `I = int(-dt)/(2t^(4))= -(1)/(2)xx(t^(-4+1))/(-4+1)+C = (1)/(6t^(3))+C` `= (1)/(6(2+(3)/(x^(2))+(1)/(x^(4)))^(3))+C" "[therefore t = 2 + (3)/(x^(2))+(1)/(x^(4))]` `= (x^(12))/(6(2x^(4)+3x^(2)+1)^(3))+C` |
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