u^2/u^3=1/u
=================
(4+u^2)/u^3=4/u^3+1/u
int(4/u^3+1/u)=int(4/u^3)+int(1/u)=int(4u^-3)+ln(u)+C=(4u^-2)/(-2)+ln(u)+C=-2(1/u^2)+ln(u)+C
Question :
integral from 1 to 2 of (4+u^2)/u^3
My work:
integreal 4/u^3+ integral u^2/u^3
4integral 1/u^3 + integral (u^2)(1/u^3)
4 integral u^-3 + integral (u^2)(u^-3)
4 integral (u^-2)/-2 + integral (u^3/3)(u^-2/-2)
4 [[(2^-2)/-2] -[(1^-2)/-2]} +{ [(2^3/3)(2^-2/-2)]-[(1^3/3)(1^-2/-2)]}
Now im stuck because i want to make my answer clean in a fraction. i can solve it usiing decimals but how would i re write it soo i end up dealing with fractions?