Question

\[\int_{2}^{3} \frac{u^3+1}{u^3-u^2} du\]

Answer 1

\[1+ \frac{u^2-1}{u^3- u^2}\]

Answer 2

\[\int_{2}^{3} \frac{u^3+1}{u^2(u-1)} du\]

Answer 3

\[1+\frac{u^2+1}{u^3-u^2}\]I think

Answer 4

yep it's plus 1

Answer 5

\[\int_{2}^{3} 1 du +\int_{2}^{3}\frac{u^2+1}{u^3-u^2} du\]

Answer 6

so we can do partial fractions\[1+\frac{u^2+1}{u^2(u-1)}=1+\frac Au+\frac B{u^2}+\frac C{u-1}\]

Answer 7

can we leave the 1+ out while doing partial fractions?

Answer 8

yes

Answer 9

\[\frac{u^2+1}{u^2(u-1)}=\frac Au+\frac B{u^2}+\frac C{u-1}\]\[Au(u-1)+B(u-1)+Cu^2=u^2+1\]

Answer 10

\[\frac{u^2+1}{u^2(u-1)}=\frac Au+\frac B{u^2}+\frac C{u-1}\]\[Au(u-1)+B(u-1)+Cu^2=u^2+1\]\[(A+C)u^2+(-A+B)u-B=u^2+1\]so\[B=-1\]which means\[A=-1\]which means\[C=2\]

Answer 11

yep

Answer 12

\[-\frac {1}{u}-\frac {1}{u^2}+\frac 2{u-1}\]
\[-\int_{2}^{3} \frac {1}{u}du-\int_{2}^{3} \frac {1}{u^2}du+\int_{2}^{3} \frac 2{u-1}du\]
\[-ln| u| +\frac1u+2ln|u-1|\] evaluated from u=2 to u=3

Answer 13

don't forget to pout the +1 back in before you integrate!

Answer 14

put*

Answer 15

oowww...yes thank you. I will pout that back in

Answer 16

Thanks Turing!

Answer 17

welcome!