Question

Inegrate (x^2)/(x^4 +1)dx

Answer 1

\[\int\limits_{}^{}x ^{2}dx/\left( x ^{4}+1 \right)\]

Answer 2

- To integrate this we use quetient rule:
=[2x(x^4)-4x^3(x^2)]/(x^4)^2
=[2x^5-4x^5]/x^6
=-2x^5/x^6

Answer 3

@WONDEMU seems you differentiated it instead of integrating,,that too wrongly! :D

Answer 4

dude what is that guy doing, I specifically said to integrate

Answer 5

any of you guys need a hint

Answer 6

Trig sub might work; \(x^2=tan\theta\) => \(2xdx=sec^2\theta\)

Answer 7

∫ x^2/(x^4 +1) dx = 1/2 ∫ (2x^2) /(x^4+1) dx

= 1/2 ∫ (x^2+1 +X^2 -1 ) /(x^4 +1) dx

= 1/2 ∫ (x^2 +1 ) /(x^4 + 1) dx + ∫ (x^2 -1 ) /(x^4 +1 ) dx

= 1/2 [ A+B]

A = ∫ (x^2 +1 ) /(x^4 + 1) dx = ∫ (1 +1/x^2 ) /(x^2 + 1/x^2) dx dividing by x^2

for solving A we can put u=x-(1/x) then we have
1 + 1/x^2 = du

also x^2 +1/x^2 = x^2 +1/x^2 -2 +2 = (x -1/x)^2 +2 = u^2 +2

A = ∫ du /(u^2 +2) = 1/√2 arctan ( u /√2) = 1/√2 arctan( (x^2 -1) /(√2x) )

B = ∫ (x^2 -1 ) /(x^4 +1 ) dx

∫ (1 -1/x^2 ) /(x^2 +1/x^2 ) dx

for solving B like A we put u=x+(1/x)
hence
1 - 1/x^2 = du
also x^2 +1/x^2 = X^2+ 1/x^2 +2 -2 = (x +1/x)^2 -2 = u^2 -2

∫ du /(u^2 -2 ) = 1/2√2 ln ((u - √2) /(u +√2) )

= 1/2√2 ln [ ( x +1/x -√2) / (x+1/x +√2) ]

= 1/2√2 ln [ (x^2 -√2 x +1 ) /(x^2 +√2 x+1) ]

so the final answer is 1/2 [A+B]+c = 1/2√2 arctan( (x^2 -1) /(√2x) ) + 1/4√2 ln [ (x^2 -√2 x +1 ) /(x^2 +√2 x+1) ]+c c is a constant

Answer 8

leme try a very indifferent method :
split denominator as (x^2 +i)(x^2 -i) where i is sqrt(-1)
now multiply divide num by 2
so we have 1/2 [2x^2/(x^2 +i)(x^2 -i) ] dx
or 1/2[(x^2 + x^2) /(x^2 +i)(x^2 -i)]dx
or 1/2[(x^2 + i + x^2 -i) /(x^2 +i)(x^2 -i)]dx
or 1/2[1/(x^2 +i) + 1/(x^2 -i)]dx
now integral of 1/(x^2 -a^2) = 1/2a (log(x-a) - log(x+a)) +C..where a is const..
also integral of 1/(x^2 +a^2) = 1/a tan^-1 (x/a) +C
same thing goes here..
let m^2 =i
so we have :
1/2 [1/2m (log(x-m)/(x+m)) + 1/m (tan^-1 (x/m) ] +C
now we just have to simplify
x-m/x+m = (x- m)^2 /(x^2 -i) = x^2 +i -2xm /(x^2 -i) = (x^2 +i -2xm)(x^2 +i) /(x^4 -1) = (x^4 + 2x^2 i -2x^3 m -1 -2xim ) /(x^4 -1)

also x/m = xm/i = -xmi

Answer 9

1/m = m/i = -mi
so we finally have :
1/2 [-mi/2 log ((x^4 + 2x^2 i -2x^3 m -1 -2xim )/(x^4 -1)) -mi (tan^-1 (-mix) ] +C

some knowledge of hyperbolic eqns might be able to bring whole into real terms..hmm

Answer 10

or even converting to exponential might help..

Answer 11

This might work as well..
\[x^2=\tan\theta => 2xdx=\sec^2d\theta \]
\[dx = \frac{\sec^{2}\theta}{2x} d\theta => \frac{\sec^{2}\theta}{2\sqrt{tanx}} d\theta \]
\[\int\limits\frac{x^{2}}{x^{4}+1} dx => \int\limits\frac{\tan\theta}{(1+\tan\theta)^{2}} *\frac{\sec^{2}\theta}{2\sqrt{\tan\theta}} \]

Answer 12

You are somewhat making progress mini_x3

Answer 13

Although you might want to recheck your last step

Answer 14

whats wrong/

Answer 15

well if you take x^2 =tan theta will x^4 + 1 be = (1+tan theta)^2 or 1+tan^2 theta

Answer 16

lawl. i made a typo sorry
\[\int\limits\frac{x^{2}}{x^{4}+1} dx => \int\limits\frac{\tan\theta}{1+(\tan\theta)^{2}} *\frac{\sec^{2}\theta}{2\sqrt{\tan\theta}} \]

Answer 17

now at least it might seem a little easier once you turn 1 +tan^2 theta into sec^2 theta

Answer 18

Yes; you can try it. (:

After than manipulate it; then two trig subs I think..

Answer 19

that*

Answer 20

Now here is the greatest problem it comes out to be:\[\int\limits_{}^{}\sqrt{\tan \theta} d \theta \]

Answer 21

with a 1/2 outside

Answer 22

dont simplify too much!!!

Answer 23

so basically what I am asking is what is the integral of root over tan theta

Answer 24

First; dont simplify it too much..
convert tanx to sinx/cosx then \(u=sinx-cosx\)

Answer 25

wait what is the integrand you are getting

Answer 26

@apoorvk: can take over :p

Answer 27

dude is there anyway to see the document it is very messy