How to integrate (x^2) / (1+x^4)?

Question

How to integrate (x^2) / (1+x^4)?

Answer 1

x^2/ x^4 + 1

P(x) = x^2 - 1

don't know if that's what u wanted...

Answer 2

u put in and u take it out

Answer 3

substitute x^2 = u , now you will have to integrate this function

integrate u / (1 + u^2) du

and

x^2 = u

2x dx = 1 du

2x = du / dx

so you must integrate u / (1 + u^2) du , where du = 2x * dx

Answer 4

very tough integral.....

∫ [x^2 / (1 + x^4)] dx =

since the denominator is not factorable, you have to rearrange it as follows:

add 2x² - 2x² to the denominator:

∫ [x^2 / (1 + x^4)] dx =

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

thus, completing the square:

∫ {x^2 / [(1 + x^4 + 2x^2) - 2x^2]} dx =

now, viewing it as a difference between two squares, you can factor it as:

∫ {x^2 / [(1 + x^2)^2 - (x√2)^2]} dx =

∫ x^2 dx / {[(1+ x^2) + x√2][(1 + x^2) - x√2]} =

∫ {x^2 / [(1 + x^2 + x√2)(1 + x^2 - x√2)]} dx =

that can be decomposed into partial fractions as:

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

(Cx + D)/(1+ x^2 - x√2) →

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

(Cx + D)(1 + x^2 + x√2)] / [(1 + x^2 + x√2)(1 + x^2 - x√2)]

thus, equating the numerators, you get:

x^2 = [(Ax + B)(1 + x^2 - x√2) + (Cx + D)(1 + x^2 + x√2)] →

x^2 = Ax + Ax^3 - Ax^2√2 + B + Bx^2 - Bx√2 + Cx + Cx^3 + Cx^2√2 + D + Dx^2 + Dx√2 →

x^2 = (A + C)x^3 + (- A√2 + B + C√2 + D)x^2 + (A - B√2 + C + D√2)x + (B + D)

yielding the system:

| A + C = 0 →

| - A√2 + B + C√2 + D = 1 → - A√2 + C√2 + (B + D) = 1 →

- A√2 + C√2 + 0 = 1 → - A√2 + C√2 = 1 →

| A - B√2 + C + D√2 = 0 → (A + C) - B√2 + D√2 = 0 → 0 - B√2 + D√2 = 0 →

D√2 = B√2 → D = B = 0

| B + D = 0

| A + C = 0 → A = - C = - 1/(2√2)

| - (-C)√2 + C√2 = 1 → 2C√2 = 1 → C = 1/(2√2)

| B = 0

| D = 0

therefore you get (see above):

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

(Cx + D)/(1+ x^2 - x√2) →

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

[1/(2√2)]x/(1+ x^2 - x√2) →

your integral thus becoming:

∫ {x^2 / [(1 + x^2 + x√2)(1 + x^2 - x√2)]} dx = ∫ { {[-1/(2√2)]x /(1+ x^2 + x√2)} +

{[1/(2√2)]x /(1+ x^2 - x√2)} } dx =

breaking it up and taking the constants out,

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

divide and multiply both integrals by 2, in order to change each numerator into the derivative of the respective denominator:

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

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

then, in order to complete the derivatives, add (√2 - √2) to both numerators:

[-1/(4√2)] ∫ [(2x +√2 - √2) /(1+ x^2 + x√2)] dx +

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

break the integrals up into:

[-1/(4√2)] ∫ {[(2x +√2)/(1+ x^2 + x√2)] - [√2 /(1+ x^2 + x√2)]} dx +

[1/(4√2)] ∫ {[(2x - √2)/(1+ x^2 - x√2)] + [√2 /(1+ x^2 - x√2)]} dx =

[-1/(4√2)] ∫ [(2x +√2)/(1+ x^2 + x√2)] dx - [-1/(4√2)] ∫ [√2 /(1+ x^2 + x√2)] dx +

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

[-1/(4√2)] ∫ [d(1+ x^2 + x√2)]/ (1+ x^2 + x√2) + [1/(4√2)] (√2) ∫ [1 /(1+ x^2 + x√2)] dx +

[1/(4√2)] ∫ [d(1+ x^2 - x√2)]/ (1+ x^2 - x√2) + [1/(4√2)] (√2)∫ [1 /(1+ x^2 - x√2)] dx =

[-1/(4√2)] ln(1+ x^2 + x√2) + (1/4) ∫ [1 /(1+ x^2 + x√2)] dx +

[1/(4√2)] ln(1+ x^2 - x√2) + (1/4) ∫ [1 /(1+ x^2 - x√2)] dx =

as for the remaining integrals, being their denominators unfactorable, you have to

complete the square, trying to rearrange them into {[f(x)]^2 + 1} form, thus integrable as arctangent:

replace 1 with (1/2) + (1/2):

[-1/(4√2)] ln(x^2 + x√2 + 1) + [1/(4√2)] ln(x^2 - x√2 + 1) +

(1/4) ∫ dx /{[x^2 + x√2 + (1/2)] + (1/2)} + (1/4) ∫ dx /{[x^2 - x√2 + (1/2)] + (1/2)} dx =

[-1/(4√2)] ln(x^2 + x√2 + 1) + [1/(4√2)] ln(x^2 - x√2 + 1) +

(1/4) ∫ dx /{[x + (1/√2)]^2 + (1/2)} + (1/4) ∫ dx /{[x - (1/√2)]^2 + (1/2)} dx =

[-1/(4√2)] ln(x^2 + x√2 + 1) + [1/(4√2)] ln(x^2 - x√2 + 1) +

(1/4) ∫ dx /{[x + (1/√2)]^2 + (1/2)} + (1/4) ∫ dx /{[x - (1/√2)]^2 + (1/2)} dx =

factor out (1/2) from both denominators:

[-1/(4√2)] ln(x^2 + x√2 + 1) + [1/(4√2)] ln(x^2 - x√2 + 1) +

(1/4) ∫ dx /{(1/2){2[x + (1/√2)]^2 + 1}} + (1/4) ∫ dx /{(1/2){2[x - (1/√2)]^2 + 1}} =

[-1/(4√2)] ln(x^2 + x√2 + 1) + [1/(4√2)] ln(x^2 - x√2 + 1) +

(1/4)(2) ∫ dx /{2[x + (1/√2)]^2 + 1} + (1/4)(2) ∫ dx / {2[x - (1/√2)]^2 + 1} =

then include 2 into the squares as:

[-1/(4√2)] ln(x^2 + x√2 + 1) + [1/(4√2)] ln(x^2 - x√2 + 1) +

(1/2) ∫ dx /{ {√2[x + (1/√2)]}^2 + 1 } + (1/2) ∫ dx /{ {√2[x - (1/√2)]}^2 + 1} =

expand the bases of the squares as:

[-1/(4√2)] ln(x^2 + x√2 + 1) + [1/(4√2)] ln(x^2 - x√2 + 1) +

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

finally, in order to turn the numerators of the remaining integrals into the derivative of

(x√2 - 1), that is √2, divide and multiply by √2 as:

[-1/(4√2)] ln(x^2 + x√2 + 1) + [1/(4√2)] ln(x^2 - x√2 + 1) +

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

[-1/(4√2)] ln(x^2 + x√2 + 1) + [1/(4√2)] ln(x^2 - x√2 + 1) +

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

[-1/(4√2)] ln(x^2 + x√2 + 1) + [1/(4√2)] ln(x^2 - x√2 + 1) +

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

[1/(4√2)] [ln(x^2 - x√2 + 1) - ln(x^2 + x√2 + 1)] +

[1/(2√2)] arctan(x√2 + 1) + [1/(2√2)] arctan(x√2 - 1)] + C

in conclusion, according to log properties, you get:

∫ [x^2 / (1 + x^4)] dx =

[1/(4√2)] [ln(x^2 - x√2 + 1) /(x^2 + x√2 + 1)] + [1/(2√2)] [arctan(x√2 + 1) +

arctan(x√2 - 1)] + C

I hope it helps...

Bye!

Answer 5

∫(x^2) / (1+x^4) dx

We will use the concept of partial fractions to solve the integral. The denominator can be factored into (1 - x^2) and (1 + x^2). (1 - x^2) can be further broken down into (1 - x) and (1 + x). Thus,

(x^2) / (1 + x^4) = (Ax + B) / (1 + x^2) + C / (1 - x) + D / (1 + x)

Multiplying both sides of the equation by (1 + x ^4):

x^2 = (Ax + B)(1 - x^2) + C(1 + x^2)(1 + x) + D(1 + x^2)(1 - x)

= Ax - Ax^2 + B - Bx^2 + C(1 + x^2 + x + x^3) + D(1 + x^2 - x - x^3)

= Ax - Ax^2 + B - Bx^2 + C + Cx + Cx^2 + Cx^3 + D - Dx + Dx^2 - Dx^3

We group similar terms:

x^2 = (B + C + D) + (A + C - D)x + (-A - B + C + D)x^2 + (C - D)x^3

Here, we note that the coefficients in the right side must correspond with the ones on the left. We only have one term on the left, x^2, which has a coefficient of 1. Thus,

-A - B + C + D = 1

The rest are 0s.

(1) B + C + D = 0

(2) A + C - D = 0

(3) C - D = 0 -> C = D

(4) -A - B + C + D = 1

Now, we solve this system. Using (3) with (2), we get A = 0. Using this in (4) together with (1):

-B + C + D = 1

B + C + D = 0

2C + 2D = 1

2C + 2C = 1

4C = 1

C = 1/4 D = 1/4

B = -1/2

Thus, A = 0, B = -1/2, C = 1/4 and D = 1/4.

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

The integral becomes:

∫(x^2) / (1+x^4) dx

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

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

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

= (-1/2) Arctan x - (1/4)ln |1 - x| + (1/4) ln |1 + x| + C

And here's your integral. Whew!

Answer 6

use partial fraction, but before that .. . . you need to factor the denominator into quadratics

1 + x^4 = x^4 + 2x^2 + 1 - 2x^2 = (x^2+1)^2 - ([√2] x)^2

thus

(x^4 + 1) = (x^2 + [√2] x + 1) (x^2 - [√2] x + 1)

thus

x^2/(x^4 + 1) = A(2x + √2)/(x^2 + [√2] x + 1) + B/(x^2 + [√2] x + 1) + C(2x - √2)/(x^2 - [√2] x + 1) + D/(x^2 - [√2] x + 1)

then you need to solve for such A, B , C and D ... will continue

then

x^2 = A(2x + √2)(x^2 - [√2] x + 1) + B(x^2 - [√2] x + 1) + C (2x - √2)(x^2 + [√2] x + 1) + D (x^2 + [√2] x + 1)

= A(2x^3 - √2 x^2 + √2) + B(x^2 - [√2] x + 1) + C(2x^3 + √2 x^2 - √2) + D(x^2 + [√2] x + 1)

= (2A + 2C) x^3 + (-√2 A + B + √2 C + D) x^2 + (-√2 B + √2 D) x + (√2 A + B - √2 C + D)

thus

A + C = 0

-√2 A + B + √2 C + D = 1

B - D = 0

√2 A + B - √2 C + D = 0

A = -1/4√2 = -C

B = 1/4 = D

thus

∫ x^2/(x^4+1) dx =

-1/4√2 ∫ (2x + √2)/(x^2 + [√2] x + 1) dx

+ 1/4 ∫ 1/(x^2 + [√2] x + 1) dx

+ 1/4√2 ∫ (2x - √2)/(x^2 - [√2] x + 1) dx

+ 1/4 ∫ 1/(x^2 + [√2] x + 1) dx

to solve the second integral:

∫ 1/(x^2 + √2 x + 1) dx = ∫ 1/(x^2 + √2 x + 1/2 + 1/2) dx ... this is arctangent with u = (x+1/√2) , a = 1/√2

(a similar approach is done for the second integral)

then original integral

= -1/(4√2) ln|x^2 + √2 x + 1|  + 1/(2√2) arctan(√2 x + 1) + 1/(4√2) ln|x^2 - √2 x + 1|  + 1/(2√2) arctan(√2 x - 1) + C

Answer 7

Not sure if this is correct but this is how I would do it:

let u = x^3

du/dx = 3x^2

dx = 1/3x^2 du..subbed back in

x^2 / (1+ u.x) * 1/3x^2  du                    

x^2 cancel and 1/3 constant taken out and x = u^1/3

1/3   *  [1/ (1+ u^4/3)]

=  3/21*u^-7/3 ln (1 + u^4/3)  then sub u = x^3 back in.