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If `int(e^(4x)-1)/(e^(2x))log((e^(2x)+1)/(e^(2x-1)))dx=(t^(2))/(2)logt-(t^(2))/(4)-(u^(2))/(2)logu+(u^(2))/(4)+C,` thenA. `u=e^(x)+e^(-x)`B. `u=e^(x)-e^(-x)`C. `t=e^(x)+e^(-x)`D. `t=e^(x)-e^(-x)` |
Answer» Correct Answer - B::C `I=int{(e^(2x)-e^(-2x))ln(e^(x)+e^(-x))-(e^(2x)-e^(-2x))ln(e^(x)-e^(-x))}dx` `=int tln t dt- int u ln u du ("where t"=e^(x)+e^(-x) and u=e^(x)-e^(-x))` `=(t^(2))/(2)ln t-(t^(2))/(4)-(u^(2))/(2)ln u+(u^(2))/(4)+C` |
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