f(x) = (integral (1/t^2-9)dt, 1 to x) dt

Question

explain reasoning.

Answer 1

integral((1/(t^2-9)dt) can be written integral((A/(t-3)+B/(t+3))dt), where A and B are constants to be calculated.

A/(t-3)+B/(t+3)=(At+3A+Bt-3B)/(t^2-9). The numerator must come to 1, so A+B=0, making B=-A, because there are no t terms, and 3A-3B=1, i.e., 3A+3A=1, so A=1/6 and B=-1/6. The integral becomes integral((1/(6(t-3))-1/(6(t+3))dt)=integral(1/(6(t-3))dt)-integral(1/(6(t+3))dt)=(1/6)ln(t-3)-(1/6)ln(t+3)=(1/6)ln|(t-3)/(t+3)| between the limits t=1 to x. So f(x)=(1/6)(ln|(x-3)/(x+3)|-ln(2/4))=(1/6)ln(2|(x-3)/(x+3)|), because -ln(1/2)=ln(2) and the logs are simply added together.