How do I integrate x/(1+x^3) ?

Answer 1

The simple fact of the matter is, that this cannot be expressed in terms of elementary functions. You can't take the integral of this using u substitution, integration by parts, partial fractions, trigonometric identities, trigonometric substitution ... none of those methods will work, which is why you won't find a solution.

Just like

Integral (sin(x^2)dx), or

Integral (e^(x^2) dx)

Edit: My bad; math kp is correct.

x/(1 + x^3) = A/(1 + x) + (Bx + C)/(x^2 - x + 1)

Note that
x^2 - x + 1 = x^2 - x + (1/4) + 1 - 1/4
= [x - (1/2)]^2 + 3/4, so

x/(1 + x^3) = A/(1 + x) + (Bx + C)/( [x - (1/2)]^2 + 3/4 )

Taking the integral of A/(1 + x) is easy; the answer is just
A*ln|1 + x|.
The integral of (Bx + C)/( [x - (1/2)]^2 + 3/4 ), however, is another issue. Let's solve this integral by itself.

Integral ( (Bx + C)/( [x - (1/2)]^2 + 3/4 ) ) dx =

Let u = x - (1/2). Then
u + (1/2) = x, and
du = dx, so we have

Integral ( (B(u + 1/2) + C) / (u^2 + 3/4) ) du

Let's multiply everything by 4 to get rid of all fractions.

Integral ( (B(4u + 2) + C) / (4u^2 + 3) du )

Integral ( (4Bu + 2B + C) / (4u^2 + 3) du)

Let's separate this into two integrals:

Integral ( (4Bu) / (4u^2 + 3) ) du +
Integral ( (2B + C) / (4u^2 + 3) ) du

Let's solve these integrals separately.

Integral ( (4Bu) / (4u^2 + 3) ) du

Pull out the constant B.

B * Integral ( 4u / (4u^2 + 3) ) du

Multiply the 4u by 2; offset this by multiplying the outside by 1/2.

(1/2)B * Integral (8u / (4u^2 + 3) ) du

Now, this is an easy integral to solve.

(1/2)B * ln |4u^2 + 3|

But u = (x + (1/2)), so we have

(1/2)B * ln |4(x + 1/2) + 3|
(1/2)B * ln |4x + 2 + 3|
(1/2)B * ln |4x + 5|

Now for this integral:

Integral ( (2B + C) / (4u^2 + 3) ) du

If we pull out the constant 2B + C, we get

(2B + C) Integral ( 1 / (4u^2 + 3) ) du

We have to use trigonometric substitution on this.

Let u = sqrt(3)/2 tan(t)
du = sqrt(3)/2 sec^2(t) dt

(2B + C) Integral ( {1 / [3tan^2(t) + 3] } sqrt(3)/2 sec^2(t) } dt)

[sqrt(3)/2] (2B + C) Integral ( {1 / 3sec^2(t)} sec^2(t) dt)

[sqrt(3)/2] (2B + C) Integral ( {1/3} ) dt =

[sqrt(3)/2] (2B + C) [t/3]

But u = [sqrt(3)/2 tan(t)], so it follows that
[2/sqrt(3)]u = tan(t), and
t = tan^(-1) ([2/sqrt(3)]u), so our answer is

[sqrt(3)/2] (2B + C) [1/3] tan^(-1) ([2/sqrt(3)]u)

Which we can simplify to

[sqrt(3)/6] (2B + C) tan^(-1) ([2/sqrt(3)]u)

But u = [x + (1/2)], so we have

[sqrt(3)/6] (2B + C) tan^(-1) ( [2/sqrt(3)] (x + (1/2) )

Whew! Tall order! Let's combine everything we have now

Integral ( x / (1 + x^3) ) dx =

A*ln|1 + x| + (1/2)B * ln |4x + 5| + [sqrt(3)/6] (2B + C) tan^(-1) ( [2/sqrt(3)] (x + (1/2) )

And guess what? We're not even done, because we have to solve for A, B, and C.

x/(1 + x^3) = A/(1 + x) + (Bx + C)/(x^2 - x + 1)

But 1 + x^3 = (1 + x) (x^2 - x + 1), so

x/[(1 + x)(x^2 - x + 1)] = A/(1 + x) + (Bx + C)/(x^2 - x + 1)

Multiply both sides by (1 + x)(x^2 - x + 1),

x = A(x^2 - x + 1) + (Bx + C)(1 + x)

This is true for all x, so let x = -1. Then

-1 = A(1 + 1 + 1) + 0
-1 = A

We can derive B and C by expanding the right hand side of that equation.

x = Ax^2 - Ax + A + Bx + Bx^2 + C + Cx
x = Ax^2 + Bx^2 + Bx - Ax + Cx + A + C

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

This give us the equations
B - A + C = 1
A + C = 0

But we know the value of A; A = -1, so C = 1.
Consequently, plugging in A = -1 and C = 1 for B - A + C = 1,
B - (-1) + (1) = 1
B = 1.

Therefore, A = -1, B = 1, C = 1

That makes our final answer

-ln|1 + x| + (1/2) * ln |4x + 5| + [sqrt(3)/6] (3) tan^(-1) ( [2/sqrt(3)] (x + (1/2) )

Or, reducing this


-ln|1 + x| + (1/2) * ln |4x + 5| + [sqrt(3)/2] tan^(-1) ( [2/sqrt(3)] (x + (1/2) ) + C

Answer 2

This is not as difficult as others make it out

you can put x/(1+x^3) = A/(1+x) +(Bx+C)/(1-x+x^2)

u can now aolve for A , B and C and then integrate

Answer 3

The simplest method you can try is to find the McLaurin expansion for the given expression and integrate term by term.

OR break int partial fractions!

Answer 4

I wish I knew.

Just a suggestion... You're losing points. You lose 5 points each time you ask a question, but you receive 2 points for each one you answer. Soooo, you might want to answer at least 3 questions for each one you ask, so at least your points will go UP and not down.

Maybe once a week you can just answer several questions, or enough to keep a good supply of points on hand for the questions you intend to ask.

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