Question

I'm stuck @solomonzelman

Answer 1

\(\color{#000000}{\displaystyle\int\limits_{~}^{~} 8(y^4+4y^2+1)^2(y^3+2y)~dy}\)

the substitution is,
\(\color{#000000}{\displaystyle u=y^4+4y^2+1}\)
\(\color{#000000}{\displaystyle du=4y^3+8y~dy}\)

you might not notice, but actually do have this expression for du inside the integral.

Answer 2

\(\color{#000000}{\displaystyle\int\limits_{~}^{~} 8(y^4+4y^2+1)^2(y^3+2y)~dy}\)
\(\color{#000000}{\displaystyle\int\limits_{~}^{~} 2(y^4+4y^2+1)^2\cdot 4(y^3+2y)~dy}\)
\(\color{#0000ff}{\displaystyle\int\limits_{~}^{~} 2(y^4+4y^2+1)^2(4y^3+8y)~dy}\)

Answer 3

oh! so do I plug in du/dy for that?

Answer 4

you don't plug the du/dy, or may you meant the correct thing, but you said it abstrusely.

\(\color{#000000}{\displaystyle u=y^4+4y^2+1}\)

\(\color{#000000}{\displaystyle du/dy=4y^3+8y}\)
and here you solve for du (not for dy, as you did in the attachment)
you need to multiply both sides times the change in y, dy.
\(\color{#000000}{\displaystyle dy=(4y^3+8y)~dy}\)

Answer 5

oh, my last line should say
\(\color{#000000}{\displaystyle du=(4y^3+8y)~dy}\)

Answer 6

I don't get that step

Answer 7

Oh, wait. now I do, lol

Answer 8

ok, good. after you finish tell me what you get;]

Answer 9

yes, correct so far!

Answer 10

not integrate (with respect to u), and substitute the x back after you are done integrating

Answer 11

I meant,

noW integrate (with respect to u), and substitute the x back after you are done integrating

Answer 12

like that?

Answer 13

you forgot the power of 3 in the very last equation, and you didn't write the +C.

Answer 14

but, other than this, I think you get the concept;)

Answer 15

Oh, wow. I'm so stupid, lol. My brain getting ahead of me and forgetting stuff..sorry