How do you evaluate the definite integral #int x/(x^2-1)# from #[2,3]#? Calculus Introduction to Integration Definite and indefinite integrals 1 Answer Narad T. Oct 15, 2016 1/2ln(8/3) Explanation: #intxdx/(x^2-1)=intxdx/((x+1)(x-1))# #x/((x+1)(x-1))=A/(x+1)+B/(x-1)# #x=A(x-1)+B(x+1)# #x=1 =>1=0+2B ; B=1/2# #x=-1 =>-1=0-2A ; A=1/2# #I=intxdx/(x^2-1)=1/2(int1dx/(x+1)+int1dx/(x-1))=1/2(ln(x+1)+ln(x-1))=1/2(ln(x+1)(x-1)# #x=3 ; I=1/2Ln8# #x=2 ; I=1/2Ln3# #I=1/2(ln8-ln3)=1/2ln(8/3)# Answer link Related questions What is the difference between definite and indefinite integrals? What is the integral of #ln(7x)#? Is f(x)=x^3 the only possible antiderivative of f(x)=3x^2? If not, why not? How do you find the integral of #x^2-6x+5# from the interval [0,3]? What is a double integral? What is an iterated integral? How do you evaluate the integral #1/(sqrt(49-x^2))# from 0 to #7sqrt(3/2)#? How do you integrate #f(x)=intsin(e^t)dt# between 4 to #x^2#? How do you determine the indefinite integrals? How do you integrate #x^2sqrt(x^(4)+5)#? See all questions in Definite and indefinite integrals Impact of this question 1194 views around the world You can reuse this answer Creative Commons License