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`(e^(2x)-1)/(e^(2x)+1)` |
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Answer» `int(e^(2x-1)/(e^(2x)+1))dx" माना "e^(x)+e^(-x)=t` `=int(e^(x)-x^(-x))/(e^(x)+e^(-x))dx=int(1)/(t)dt" "rArr (e^(x)-e^(-x))dx=dt` `=log|t|+c=log|e^(x)+e^(-x)|+c` |
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