Question

http://learn.flvs.net/educator/temp/atucker69/2579/ChapterEight/8.01.pdf help on #6 please

Answer 1

#6 page 522? that's a u-substitution. set u = the denominator

Answer 2

yes that one

Answer 3

i dont get what you mean by that

Answer 4

the integrand is \(\large \frac{2t-1}{t^2-t+2} \)...
set u= the denominator of this fraction.

Answer 5

then you need to calculate du = ???

Answer 6

\[\int\limits \frac{2t - 1}{t^2-t+2}\mathrm{d}t\]

let \[u = t^2 - t + 2\]
\[\frac{\mathrm{d}u}{\mathrm{d}t} =2t -1\]
\[\mathrm{d}u = (2t - 1) \mathrm{d}t\]

Can you go on from here?
If this looks unfamiliar try watching this video (should make u-substution clear): http://www.youtube.com/watch?v=qclrs-1rpKI

Answer 7

@Meepi i got (-1/t^2+t=+2)x

Answer 8

i dont know if its right tho

Answer 9

Substitute u = t^2 - t + 2 and du = (2t - 1) dt into the original problem
\[\int\limits \frac{2t - 1}{t^2 - t + 2} dt = \int\limits \frac{1}{u} du \] with u = t^2 - t + 2
Solve the integral and substitute u back in