Question

integral from 1 to 2 (t+1/t)^2 dx

Answer 1

square first

Answer 2

\[(t+\frac{1}{t})^2=t^2+2+\frac{1}{t^2}\] integrate term by term

Answer 3

tehen seperate

Answer 4

thanks so much

Answer 5

o wait i was thinking

\[(\frac {t+1}{t})^2\]

Answer 6

how do I integrate 1/t^2

Answer 7

@satellite73: can you please help me on other problems too, I have calc exam this Monday and it's holiday weekend therefore school is closed, no tutor :((

Answer 8

\[\frac{1}{t^2}=t^{-2}\]

Answer 9

then just use the simple add one and divide by that number rule

Answer 10

\[\int t^n=\frac{t^{n+1}}{n+1}\]

Answer 11

will that be -t/-1 ?

Answer 12

\[\int t^{-2}=\frac{t^{-2+1}}{-2+1}=\frac{t^{-1}}{-1}=-t^{-1}=\frac{-1}{t}\]

Answer 13

if you ever feel like you've messed up just take the derivative of what you get

\[\frac{d}{dx}\frac{-1}{t}=-1(-1)\frac{1}{t^2}=\frac{1}{t^2}=t^{-2}\]

Answer 14

oh ic thanks. can you please help me on some other problems as well?