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`int[sin(logx)+cos(logx)]dx`A. `x log (logx)+C`B. `sin(logx)+C`C. `cos(logx)+C`D. `x sin (logx)+C` |
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Answer» Correct Answer - D `int[sin(logx)+cos(logx)]dx` `=intsin(logx)dx+intcos(logx)dx` `=x sin (logx)-int(xcos (logx))/(x)dx+intcos(logx)dx+C` `=x sin (logx)+C` |
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