Question

integral of (3t-2)/(t+1)

Answer 1

\[\Large \int \left(\frac{3t-2}{t+1}\right) dt\]


\[\Large \int \left(\frac{3(u-1)-2}{u}\right) du\]


\[\Large \int \left(\frac{3u-3-2}{u}\right) du\]


\[\Large \int \left(\frac{3u-5}{u}\right) du\]


\[\Large \int \left(\frac{3u}{u}-\frac{5}{u}\right) du\]


\[\Large \int \left(3-\frac{5}{u}\right) du\]


I'll let you finish up

Answer 2

I let u = t+1, so t = u-1

Deriving both sides of u = t+1 with respect to t gives

du/dt = 1

du = dt

Answer 3

this helps a lot man i appreciate it. missed class when we learned this...

Answer 4

then you just substitute back in after you have the final integral?