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?

2. 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

3. Originally Posted by Also sprach Zarathustra
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
subbing in [1,2] into -2(1/u^2)+ln(u) i got
=[[-2(1/2^2)]-[-2(1/1^2)]] +[ln(2)- ln(1)]
which i then got 1.5+ln2 which is 2.19?

It looks really ugly, did i do something wrong?

4. Yes! This is the correct answer!