Question

Find the indefinite Integral

Answer 1

∫(t−2t^4)/t√t

Answer 2

I'm not really sure how to start this one. I figured u = (t - 2t^4) but I'm not sure

Answer 3

\[\int\limits \frac{\left(t-2 t^4\right) \sqrt{t}}{t} \, dt ?\]

Answer 4

\[\int\limits_{?}^{?}(t-2t ^{4})/(t \sqrt{t}) dt?\]

Answer 5

try sub \[u=\sqrt{t}\]

Answer 6

\[\sqrt{t} \] on the bottom, no addition t, I'm sorry, I'm not sure how to do that divide bar Sam, how do I do that?

Answer 7

\[d u=\frac{1}{2 \sqrt{t}}d t\]

Answer 8

\[\int\limits \frac{t-2 t^4}{t \sqrt{t}} \, dt?\]

Answer 9

\[\int\limits \frac{t-2 t^4}{\sqrt{t}} \, dt?\]

Answer 10

just multiply and separate those bad boy fractions

Answer 11

wait i don't know which one we are doing
but separate those fractions lol

Answer 12

yeah

Answer 13

-.-

Answer 14

Yeah sam that one, sorry... How do I do a dividing bar like that :(?

Answer 15

so multiply t^-1/2 into t and -2t^4?

Answer 16

yes!

Answer 17

\[\int\limits \frac{t-2 t^4}{\sqrt{t}} \, dt\]
\[\int\limits \frac{t}{\sqrt{t}}-\frac{2t^4}{\sqrt{t}} \, dt\]
\[\int\limits t^{1/2}-2t^{7/2}\, dt\]
Integrate

Answer 18

so it would be \[t^{1 - 1/2} =t ^1/2 \] and

\[-2t^\]\[-2t^{4 - 1/2}\] ? Which should be \[-2t^{4/4 - 2/4} \] = 2/4? = 1/2? Is that righty?

Answer 19

hmm 7/2....

Answer 20

4 minus half is?

Answer 21

3.5 so yeah 7/2...

Answer 22

I konfused myself :P

Answer 23

yes, so integrate \[\int\limits\limits t^{1/2}-2t^{7/2}\, dt\]

Answer 24

\[2t^{3/2}/3 - 2^{9/2}/9\] How do I do the cool divide? :(

Answer 25

Should be
\[\frac{2 t^{3/2}}{3}-\frac{4 t^{9/2}}{9}+c\]
and don't forget about the constant

Answer 26

Yeah I realized that i didn't put that but didn;'t wan to redo it all ;P ty. How do I do that division instead of '/'?

Answer 27

use \frac{numerator}{denominator}

Answer 28

Is there a specific button for that? Ty

Answer 29

no

Answer 30

Oh wells... Thanks for all the help :)