Question

ln(1+根号((1+x)/x))dx

Answer 1

为方便计算,可设 (x+1)/x=u²,则 x=1/(u²-1)；

∫ln{1+√[(x+1)/x]} dx=∫ln(1+u)d[1/(u²-1)]=[ln(1+u)]/(u²-1)-∫[1/(u²-1)]*[1/(1+u)]du

=[ln(1+u)]/(u²-1)-(1/4)∫{[1/(u-1)]+[1/(1+u)]+[2/(1+u)²]}du

=[ln(1+u)]/(u²-1)-(1/4){ln(u-1)+ln(1+u)-[2/(1+u)]}

=[ln(1+u)]/(u²-1)-(1/4){ln(u²-1)-[2/(1+u)]}

=x*ln{1+√[(x+1)/x]} + (1/4)lnx + 2/{1+√[(x+1)/x]}；