Question

can someone help me with this integral ? ∫6x^3/(2x^4+1)

Answer 1

the top is almost a derivative of the bottom ... what is the derivative of the bottom?

Answer 2

also, recall derivative of ln(u) to see why thats one of the first things you try to look at.

Answer 3

i don't get where the 3/4 comes from...

Answer 4

what is the derivative of the bottom?

Answer 5

what is the deriative of: 2x^4 + 1

Answer 6

8x^3

Answer 7

good, then we would be sitting pretty if the top was 8x^3, but it aint, so we need to use some math to transform it into an 8

Answer 8

do you agree that 8/8 = 1?

Answer 9

lol yes

Answer 10

\[\int\frac{6x^3}{2x^4+1}\]
pull out the 6 and multiply it all by 8/8, why?
\[6\int\frac{8}{8}\frac{x^3}{2x^4+1}\]
use the top 8 and pull the bottom 8 out of the way
\[\frac 68\int\frac{8x^3}{2x^4+1}\]
etc ...

Answer 11

now we integrate it up into: 6/8 ln(2x^4+1) + C

Answer 12

awesome! thank you!

Answer 13

youre welcome