How do you evaluate the definite integral #int (1+y^2)/y dy# from #[1,2]#?

Question

4941 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-12-05T09:12:24

#ln2+3/2#

#int_1^2(1+y^2)/ydy#

#=int_1^2(1/y+y^2/y)dy=int_1^2(1/y+y)dy#

#=[lny+y^2/2]_1^2#

#=[lny+y^2/2]^2-[lny+y^2/2]_1#

#=[ln2+2^2/2]-[cancel(ln1)^0+1^2/2]#

#=ln2+2-1/2#

#=ln2+3/2#