Question

hey, can someone please help me with this question
intregral = x/(3+x^4) dx

Answer 1

\[\int\limits \frac{ x }{ 3+x^{2} } dx\]

Answer 2

i am stuck at \[\frac{ 1 }{ 2 } \int\limits \frac{ 1 }{ 3+u^ 2}\]

Answer 3

forgo du

Answer 4

I assume your original question should be:\[\int\limits \frac{ x }{ 3+x^{4} } dx\]

Answer 5

and that you then used the substitution:\[u=x^2\]to end up with:\[\frac{ 1 }{ 2 } \int \frac{ 1 }{ 3+u^ 2}du\]

Answer 6

your next step should be to factor out the 3 from the denominator to get:\[\frac{1}{6} \int \frac{1}{1+\frac{u^ 2}{3}}du\]

Answer 7

finally use the substitution:\[s=\frac{u}{\sqrt{3}}\]

Answer 8

hopefully you can complete from here

Answer 9

why do i have to factor out 3?

Answer 10

to try and get it into one of the standard forms

Answer 11

where did u get u/sqrt(3)

Answer 12

I am trying to get it to the form:\[\int\frac{1}{1+x^2}dx=\arctan(x)+C\]

Answer 13

\[\frac{u^2}{3}=\left(\frac{u}{\sqrt{3}}\right)^2\]

Answer 14

does it make sense now?

Answer 15

umm kinda, so do i just use 1/3 arctan(x/3) + C

Answer 16

use the "s" substitution I gave above. The integral will ten come out in terms of "s". You then need to convert back to "x".

Answer 17

*then

Answer 18

I = 1/sqrt(3) arctan u/sqrt(3) + C

Answer 19

I = 1/2 (1/sqrt(3) arctan x/sqrt(3)) + C?

Answer 20

not quite - can you please show all your steps so that I can spot where you may have made a mistake?

Answer 21

@asnaseer its x^2 right.

Answer 22

using trig identiy tan^2x +1 = sec^2x rigjt.

Answer 23

are you stating that the final answer to this integral is \(x^2\)?

Answer 24

I = 1/2 (1/sqrt(3) arctan x^2/sqrt(3)) + C? this one

Answer 25

yes :)

although a simpler way to write that would be:\[\frac{\arctan(\frac{x^2}{\sqrt{3}})}{2\sqrt{3}}+C\]

Answer 26

oh i see thanks!

Answer 27

yw :)