InterviewSolution
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InterviewSolution
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If `int (e^x-1)/(e^x+1)dx=f(x)+C,` then f(x) is equal toA. `2log(e^(x)+1)+C`B. `log(e^(2x)-1)+C`C. `2log(e^(x)+1)-x+C`D. `log(e^(2x)+1)+C` |
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Answer» Correct Answer - c We have , `int(e^(x)-1)/(e^(x)+1)dx` `rArr = int(e^(x))/(e^(x)+1)dx-int(1)/(e^(x)+1)dx` `rArr = int(1)/(e^(x)+1)d(e^(x)+1)+int(1)/(e^(-x)+1)d(e^(-x)+1)` `rArr=log(e^(x)+1)+log(e^(-x)+1)+C` `rArr=log(e^(x)+1)+log((e^(x)+1)/(e^(x)))+C` `rArr=log(e^(x)+1)^(2)-loge^(x)+C=2log(e^(x)+1)-x+C` |
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