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प्रश्न
Evaluate:
`int{1/logx - 1/(logx)^2}dx`; (where x > 1).
मूल्यांकन
उत्तर
`int{1/log_ex - 1/(log_ex)^2}dx`
= `intdx/log_ex - int1/(log_ex)^2dx`
= `int1/log_ex int dx - int{d/dx(1/log_ex) intdx}dx - int1/(log_ex)^2dx`
= `x/log_ex + int1/(log_ex)^2 1/x * x*dx - int1/(log_ex)^2dx`
= `x/log_ex + int1/(log_ex)^2dx - intdx/(log_ex)^2`
= `x/log_ex + c`
Where ′c′ is any arbitary constant of integration.
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