InterviewSolution
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InterviewSolution
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`Ifint(x^4+1)/(x^6+1)dx=tan^(-1)f(x)-2/3tan^(-1)g(x)+C ,t h e n`both `f(x)a n dg(x)`are odd functions`f(x)`is monotonic function`f(x)=g(x)`has no real roots`int(f(x))/(g(x))dx=-1/x+3/(x^3)+c`A. both `f(x)` and `g(x)` are odd functionsB. `f(x)` is one-one functionC. `f(x)=g(x)` has no real rootsD. `int (f(x))/(g(x))dx=(1)/(x)+(3)/(x^(3))+c` |
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Answer» Correct Answer - A::C::D Let `I=int((x^(4)+1))/((x^(6)+1))dt=int((x^(2)+1)^(2)-2x^(2))/((x^(2)+1)(x^(4)-x^(2)+1))dx` `=int((x^(2)+1)dx)/((x^(4)-x^(2)+1))-2 int (x^(2)dx)/((x^(6)+1))` `=int((1+(1)/(x^(2)))dx)/((x^(2)-1+(1)/(x^(2))))-2 int (x^(2)dx)/((x^(3))^(2)+1)` In the first integral, put `x-(1)/(x)=t," i.e., " (1+(1)/(x^(2)))dx=dt` and in the second integral put `x^(3) =u, " i.e., " x^(2)dx=(du)/(3)` Then ` I=int (dt)/(1+t^(2))-(2)/(3)int(du)/(1+u^(2))=tan^(-1) t-(2)/(3)tan^(-1)u +C` `=tan^(-1)(x-(1)/(x))-(2)/(3)tan^(-1)(x^(3))+C` Here, `f(x)=x-(1)/(x) and g(x)=x^(3)` Draw the graphs of `f(x) and g(x)` . We find that `f(x)` is many-one and `f(x)=g(x)` has no real roots. `int(f(x))/(g(x))dx=int(x-(1)/(x))/(x^(3))dx=int((1)/(x^(2))-(1)/(x^(4)))dx= -(1)/(x)+(3)/(x^(3))+C` |
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