Evaluate the integral #int_1^2 dt/(8-3t)#?

1 Answer
Jim G.
Mar 15, 2017

#1/3ln(5/2)~~0.305" to 3 decimal places"#

Explanation:

Using the standard result for integrals of the form.

#color(red)(bar(ul(|color(white)(2/2)color(black)(int((f'(x))/(f(x)))dx=ln|f(x)|+c)color(white)(2/2)|)))#

#f(t)=8-3trArrf'(t)=-3#

#rArrint_1^2(dt)/(8-3t)#

#=-1/3int_1^2(-3)/(8-3t)dt#

#=-1/3[ln(8-3t)]_1^2#

#-1/3(ln2-ln5)#

#1/3(ln5-ln2)#

#=1/3ln(5/2)~~0.305" to 3 decimal places"#

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