Evaluate the integral log(x+1)-logx dx/x(x+1).

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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