Find the definite integral #int_0^1dx/(2x-3)#? Calculus Introduction to Integration Sigma Notation 1 Answer Shwetank Mauria Oct 12, 2016 #int_0^1dx/(2x-3)=-0.5493# Explanation: Let #u=2x-3#, hence #du=2dx# and #intdx/(2x-3)=1/2int(du)/u=1/2lnu=1/2ln|2x-3|# Hence, #int_0^1dx/(2x-3)# = #1/2ln|2x-3|_0^1# = #1/2ln|-1|-1/2ln|-3|# = #1/2ln1-1/2ln3# = #0-0.5493# = #-0.5493# Answer link Related questions How does sigma notation work? How do you use sigma notation to represent the series #1/2+1/4+1/8+…#? Use summation notation to express the sum? What is sigma notation for an arithmetic series with first term #a# and common difference #d# ? How do you evaluate the sum represented by #sum_(n=1)^5n/(2n+1)# ? How do you evaluate the sum represented by #sum_(n=1)^(8)1/(n+1)# ? How do you evaluate the sum represented by #sum_(n=1)^(10)n^2# ? What is sigma notation for a geometric series with first term #a# and common ratio #r# ? What is the value of #1/n sum_{k=1}^n e^{k/n}# ? Question #07873 See all questions in Sigma Notation Impact of this question 1940 views around the world You can reuse this answer Creative Commons License