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a) log(e^x - 1)
b) log (1 - e^x )
c) log(e^-x - 1)
d) log (1 - e^-x)
Correct answer is option 'D'. Can you explain this answer?
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a) log(e^x - 1)b) log (1 - e^x )c) log(e^-x - 1)d) log (1 - e^-x)Corre...
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a) log(e^x - 1)b) log (1 - e^x )c) log(e^-x - 1)d) log (1 - e^-x)Corre...
Substitute u=e^x. so du= e^-x dx put all thin in eqn and apply partial fraction expression becomes. 1/(u-1) - 1/u integrate both as ln(e^x-1/e^x) So ANSWER is B) log(|1 - e^-x|)
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a) log(e^x - 1)b) log (1 - e^x )c) log(e^-x - 1)d) log (1 - e^-x)Corre...
D
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a) log(e^x - 1)b) log (1 - e^x )c) log(e^-x - 1)d) log (1 - e^-x)Correct answer is option 'D'. Can you explain this answer?
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