Find the definite integral #int_0^1dx/(2x-3)#?

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Answer 1 · 2016-10-12T15:09:50

#int_0^1dx/(2x-3)=-0.5493#

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#