Question

Can someone help me with this fractions question?
How do you get the left side to equal the right side?

Answer 1

Ive attached a screen shot. Thanks!

Answer 2

hint:
we can write:
\[\large {x^6} + 3{x^4} + 3{x^2} + 1 = {\left( {{x^2} + 1} \right)^3}\]

Answer 3

Yep... I can see that, but would the expression be \[-\frac{ x }{ (x^2+1)^3 }\] if you subtract?

Answer 4

Sorry not cube just \[-\frac{ x }{ (x^2+1) }\]

Answer 5

what is the least common multiple between
\[{\left( {{x^2} + 1} \right)^3}\]
and
\[{\left( {{x^2} + 1} \right)^2}\]?

Answer 6

\[(x^2+1)^5\] ?

Answer 7

no, it is
\[{\left( {{x^2} + 1} \right)^3}\]

Answer 8

so the right side can be rewrtten as below:
\[\frac{x}{{{{\left( {{x^2} + 1} \right)}^2}}} - \frac{x}{{{{\left( {{x^2} + 1} \right)}^3}}} = \frac{{x\left( {{x^2} + 1} \right) - x \cdot 1}}{{{{\left( {{x^2} + 1} \right)}^3}}} = ...?\]

Answer 9

OK! How did you make the numerator of the right side like that??? I get the answer now! Just thats confusing, how does x^3 become that whole thing :S

Answer 10

Ohh

Answer 11

My brain shuts down after studying for exams...

Answer 12

I get it now

Answer 13

I hope they dont ask this on the test but if they do ill remember this haha thanks

Answer 14

the left side is:
\[\frac{{{x^3}}}{{{{\left( {{x^2} + 1} \right)}^3}}}\]
since:
\[{x^6} + 3{x^4} + 3{x^2} + 1 = {\left( {{x^2} + 1} \right)^3}\]

Answer 15

thanks :)

Answer 16

Yep I got it :)
This is just the first step to the problem, I actually have to integrate that lol

Answer 17

Sigh

Answer 18

it is very simple:
in each integral we have to change variable, like below:

\[\large {x^2} + 1 = t\]
where t is the new variable

Answer 19

Yeah Im going to use \[\frac{ x }{ (x^2+1)^2 } -\frac{x}{(x^2+1)^3}\] and split them into separate integrals and try doing u-substitution

Answer 20

yes!
we have:
\[\Large 2x\;dx = dt\]

Answer 21

Ohh God, I solved it :D

Answer 22

Thanks for all the help @Michele_Laino

Answer 23

Now onto the next one ><

Answer 24

Thanks @OdinMW