Grade 11Differential Calculus
integration of log(x+1)-logx dx/x(x+1) please evaluate
integration of log(x+1)-logx dx/x(x+1) please evaluate
1 Answer
Arun
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
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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