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Question
Evaluate : `intlogx/(1+logx)^2dx`
Solution
Problem:
`intlogx/(1+logx)^2dx`
adding and substracting 1 from numerator
`int (1-1+logx)/(1+logx)^2dx`
`int (1+logx)/(1+logx)^2dx-int(1)/(1+logx)^2 dx`
`int 1/(1+logx)dx-int(1)/(1+logx)^2 dx`
For the integral
` int 1/(1+logx)dx`
integrate by parts within the sum: ∫fg'=fg−∫f'g
`f= 1/(1+logx)dx, g'=1`
`f'=-(1)/(1+logx)^2, g=x`
`=-int(1)/(1+logx)^2 dx-int -1/(1+logx)^2dx+x/(log(x)+1)`
`=x/(log(x)+1)`
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