Integration Question (1 Viewer)

[cutemouse]

Integrate dx/(x²(1+x²))

Thanks to anyone who can help me..

 

[ibrian]

I = int (dx/(x²+x⁴)
= (1/2x+4x³)int(1/2x+4x³dx/(x²+x⁴)
= (1/2x+4x³)(log[x²+x⁴]) + c

 

[untouchablecuz]

let x = tanu

dx/du = (secu)^2

S dx/((x^2)(1+x^2))

S [(secu)^2du]/[(tan^2u)(1+tan^2u)]

S [du(secu)^2]/[(tan^2u)(secu)^2)]

S [du]/[tan^2u]

S [cot^2u]du

S [cosec^2u -1]du

S [(sec(pi/2 - u))^2 - 1]du

-tan(pi/2 - u) - u + C

-cot(u) - u + C

now just convert it to algebraic form using the triangle formed from tanu = x/1

you could also do it using partial fractions, its MUCH easier

1/(x^2)(x^2+1) = 1/x^2 - 1/(x^2+1) (by trial and error, easier than using the long winded method)

 

[Dumbledore]

use partial fractions; i don't have latex

1/((x^2)(1+x^2) = a/x^2+b/x+(cx+d)/(1+x^2)
a= 1/(1+0) = 1 [heaviside cover method]
(1+x^2) +bx(1+x^2) + (cx+d)x^2 = 1
substitute mutual value: x=i
(1+i^2) + bi(1+i^2) + (ci+d)(i^2) = 1
(i^2=-1 so 1+i^2 = 0)
ci+d = -1
c = 0, d = -1 (equating complex and real)
equating x of degree 1; b=0
therefore integral[dx/((x^2)(1+x^2))] = integral[1/x^2-1/(1+x^2)]dx
= -1/x - arctan(x) + c

 

[shuning]

回复: Re: Integration Question

use partial fractions; i don't have latex

1/((x^2)(1+x^2) = a/x^2+b/x+(cx+d)/(1+x^2)
a= 1/(1+0) = 1 [heaviside cover method]
(1+x^2) +bx(1+x^2) + (cx+d)x^2 = 1
substitute mutual value: x=i
(1+i^2) + bi(1+i^2) + (ci+d)(i^2) = 1
(i^2=-1 so 1+i^2 = 0)
ci+d = -1
c = 0, d = -1 (equating complex and real)
equating x of degree 1; b=0
therefore integral[dx/((x^2)(1+x^2))] = integral[1/x^2-1/(1+x^2)]dx
= -1/x - arctan(x) + c

think u made a mistake there ^^

 

[Trebla]

Integrate dx/(x²(1+x²))

Thanks to anyone who can help me..

 

[Trebla]

I = int (dx/(x²+x⁴)
= (1/2x+4x³)int(1/2x+4x³dx/(x²+x⁴)
= (1/2x+4x³)(log[x²+x⁴]) + c

Um...you can't take variables out of the integral. You can only do that to constants. It's a bit like saying ∫ x dx = x ∫ dx = x² + c which is clearly FALSE as ∫ x dx = x²/2 + c.

 

[Dumbledore]

Re: 回复: Re: Integration Question

think u made a mistake there ^^

whats wrong with it?



neat@ treblas method

 

[shuning]

回复: Re: 回复: Re: Integration Question

see the part i highlighted? if the bottom is x^2 the top has to be bx+ c LOL. but looks like the bottom is x. either way u made some stupid mistake there xD ( sry for my shitty standard english expression im sure some1 will clean it up 4 me T.T )

and yea Treblas's method is god ^^

 

[Dumbledore]

Re: 回复: Re: 回复: Re: Integration Question

see the part i highlighted? if the bottom is x^2 the top has to be bx+ c LOL. but looks like the bottom is x. either way u made some stupid mistake there xD ( sry for my shitty standard english expression im sure some1 will clean it up 4 me T.T )

and yea Treblas's method is god ^^

u only put cx+d if the denominator is irreducable
example (cx+d)/x^2 = c/x + d/x^2 and i put a a/x^2 and b/x so it works out

 

[shuning]

回复: Re: 回复: Re: 回复: Re: Integration Question

its x^2 LOl the power is greater than 1...

 

[Dumbledore]

there is a separate rule for linear factors raised to a power

e.g. a/x^4+b/x^3+c/x^2+d/x can be rewriten as (a+bx+cx^2+dx^3)/x^4
they still cater for the same quantity of variables but are written differently

what u expected was (ax+b)/x^2 but i rewrote this as a/x+b/x^2

 

[shuning]

回复: Re: Integration Question

o didnt see that LOL. learn LaTeX xD
i thought u wrote a/x^2 xD

 

[ibrian]

Um...you can't take variables out of the integral. You can only do that to constants. It's a bit like saying ∫ x dx = x ∫ dx = x² + c which is clearly FALSE as ∫ x dx = x²/2 + c.

oo whoops big mistake LOl thanks