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integration of log(x+1)-logx dx/x(x+1) please evaluate

integration of log(x+1)-logx dx/x(x+1) please evaluate

Grade:11

1 Answers

Arun
25768 Points
2 years ago
I = [log (x+1) -log (x)]/[x (x+1)] dx 

let [log (x+1) -log (x)] = t 
》 [1/(x+1)-1/x ] dx = dt 
》[(x-x-1)/(x (x+1)] dx = dt 
》 -1/(x (x+1)) dx = dt 
》 1/(x (x+1)dx = -dt 
I = -t dt 

= -t^2/2 + c 
= -[log (x+1) - logx ]^2 + c 
= - [log (x+1)/x]^2 + c

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