Question

Calc integral question...I just cant figure this one out to save my life:

Answer 1

What's troubling you?

Answer 2

Since 1-8t=u
u=1-8t
du=8dt

1/8 Integral of 1/u

Answer 3

Hmm i got a really long answer, i dont think its right. what did you guys get?

Answer 4

u = 1-8t

So du/dt = -8, which means that du = -8dt

So basically, dt = -du/8


Now perform the substitutions to get


\[\int\frac{dt}{(1-8t)^6}=-\frac{1}{8}\int\frac{du}{u^6}\]

Answer 5

wait so what did you get as ur answer?

Answer 6

\[-\frac{1}{8}\int\frac{du}{u^6}=-\frac{1}{8}\int u^{-6}du=-\frac{1}{8}\cdot \frac{u^{-5}}{-5}+C = \frac{1}{40(1-8t)^5}+C\]