∫dx/[(x+a)^2-b^2]=(1/2b)∫dx[1/(x+a-b)-1/(x+a+b)]=(1/2b){∫d(x+a-b)[1/(x+a-b)]-∫d(x+a+b)[1/(x+a+b)]}=(1/2b)[ln(x+a-b)-ln(x+a+b)]=(1/2b)ln[(x+a-b)/(x+a+b)]