I = [log (x+1) -log (x)]/[x (x+1)] dx
misalkan [log (x+1) -log (x)] = t
[1/(x+1)-1/x ] dx = dt
[(xx-1)/(x (x+1)] dx = dt
-1/(x (x+1)) dx = dt
1/(x (x+1)dx = -dt
saya = -t dt
= -t^2/2 + c
= -[log (x+1) – logx ]^2 + c
= – [log (x+1)/x]^2 + c
I = [log (x+1) -log (x)]/[x (x+1)] dx
misalkan [log (x+1) -log (x)] = t
[1/(x+1)-1/x ] dx = dt
[(xx-1)/(x (x+1)] dx = dt
-1/(x (x+1)) dx = dt
1/(x (x+1)dx = -dt
saya = -t dt
= -t^2/2 + c
= -[log (x+1) – logx ]^2 + c
= – [log (x+1)/x]^2 + c