Evaluate the Integral ?

Question

#int_1^2((v^3-3v^6)/v^4)dv#
I got to the point where I used the properties of integrals and my answer was #ln2 +5#. Where F(b)-F(a). The answer was #ln2+7#. Any ideas where I went wrong. I don't see where they are getting 7 from.

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Answer 1 · 2018-08-11T18:52:22

#ln2-7#.

#int_1^2((v^3-3v^6)/v^4)dv#,

#=int_1^2(v^3/v^4-3v^6/v^4)dv#,

#=int_1^2(1/v-3v^2)dv#,

#=[ln|v|-3*v^(2+1)/(2+1)]_1^2#,

#=[ln|v|-v^3]_1^2#,

#=[{ln|2|-2^3}-{ln|1|-1^3}]#,

#=ln2-8-0+1#.

#rArr int_1^2((v^3-3v^6)/v^4)dv=ln2-7#.