How do I evaluate this integral: 1/(x^3+x+2) dx?

Question

I found the partial fraction to be (x^2+1/x)(x+1/x) but everything else does seem to go according to plan

Answer 1

∫ [1 / (x³ + x + 2)] dx =

before using partial fractions, you have to factor the denominator properly;

thus add (x² - x² + x - x) (=0) to the denominator:

∫ [1 / (x³ + x + 2 + x² - x² + x - x)] dx =

and rearrange the terms as:

∫ [1 / (x³ + 2x + 2 + x² - x² - x)] dx =

∫ [1 / (x³ - x² + 2x + x² - x + 2)] dx =

group the terms as:

∫ {1 / [(x³ - x² + 2x) + (x² - x + 2)]} dx =

factor out x from the first group:

∫ {1 / [x(x²  - x + 2) + (x² - x + 2)]} dx =

then, factoring out the common term (x² - x + 2), you get:

∫ {1 / [(x² - x + 2)(x + 1)]} dx =

being the denominator completely factored, set partial fraction decomposition:

1 / [(x + 1)(x² - x + 2)] = A/(x + 1) + (Bx + C)/(x² - x + 2) (#) →

1 /[(x + 1)(x² - x + 2)] = [A(x² - x + 2) + (Bx + C)(x + 1)] /[(x + 1)(x² - x + 2)] →

then, equating the numerators you get:

1 = [A(x² - x + 2) + (Bx + C)(x + 1)] →

1 = Ax² - Ax + 2A + Bx² + Bx + Cx + C →

1 = (A + B)x² + (- A + B + C)x + (2A + C) →

yielding:

| A + B = 0

| - A + B + C = 0

| 2A + C = 1

| A = - B

| - (- B) + B + C = 0

| 2(- B) + C = 1

| A = - B

| B + B + C = 0

| - 2B + C = 1

| A = - B

| 2B + C = 0

| - 2B + C = 1

| A = - B

| C = - 2B

| - 2B + (- 2B) = 1

| A = - B

| C = - 2B

| - 2B - 2B = 1

| A = - B

| C = - 2B

| - 4B = 1

| A = - B = (1/4)

| C = - 2B = - 2(-1/4) = (1/2)

| B = (-1/4)

therefore (see (#) above):

1 / [(x + 1)(x² - x + 2)] = A/(x + 1) + (Bx + C)/(x² - x + 2) →

1 / [(x + 1)(x² - x + 2)] = (1/4)/(x + 1) + [(-1/4)x + (1/2)]/(x² - x + 2) →

factoring out (1/4) and (-1/4),

1 / [(x + 1)(x² - x + 2)] = (1/4)[1/(x + 1)] + (-1/4) [(x - 2)/(x² - x + 2)]

the given integral becoming:

∫ {1 / [(x² - x + 2)(x + 1)]} dx = ∫ {(1/4)[1/(x + 1)] + (-1/4)[(x - 2)/(x² - x + 2)]} dx =

breaking it up and taking out the constants,

(1/4) ∫ [1/(x + 1)] dx - (1/4) ∫ [(x - 2)/(x² - x + 2)] dx =

(1/4) ln |x + 1| - (1/4) ∫ [(x - 2)/(x² - x + 2)] dx =

as for the remaining integral, attempting to turn the numerator into the derivative

of the denominator, dvide and multiply by 2:

(1/4) ln |x + 1| - (1/4)(1/2) ∫ [2(x - 2)/(x² - x + 2)] dx =

(1/4) ln |x + 1| - (1/8) ∫ [(2x - 4)/(x² - x + 2)] dx =

in order to complete the derivative, replace - 4 with -1 - 3:

(1/4) ln |x + 1| - (1/8) ∫ [(2x - 1 - 3)/(x² - x + 2)] dx =

break it up into:

(1/4) ln |x + 1| - (1/8) ∫ {[(2x - 1)/(x² - x + 2)] - [3 /(x² - x + 2)]} dx =

(1/4) ln |x + 1| - (1/8) ∫ [(2x - 1)/(x² - x + 2)] dx + (1/8) ∫ [3 /(x² - x + 2)] dx =

(1/4) ln |x + 1| - (1/8) ∫ [d(x² - x + 2)] /(x² - x + 2) + (3/8) ∫ [1 /(x² - x + 2)] dx =

(1/4) ln |x + 1| - (1/8) ln (x² - x + 2) + (3/8) ∫ [1 /(x² - x + 2)] dx =

as, for the last integral, being the denominator unfactorable, complete the square as:

(1/4) ln |x + 1| - (1/8) ln (x² - x + 2) + (3/8) ∫ dx /[x² - x + 2 + (1/4) - (1/4)] =

(1/4) ln |x + 1| - (1/8) ln (x² - x + 2) + (3/8) ∫ dx /{[x² - x + (1/4)] + [2 - (1/4)]} =

(1/4) ln |x + 1| - (1/8) ln (x² - x + 2) + (3/8) ∫ dx / {[x - (1/2)]² + (7/4)} =

in order to rearrange the denominator into {[f(x)]²+ 1} form, factor out (7/4):

(1/4) ln |x + 1| - (1/8) ln (x² - x + 2) + (3/8) ∫ dx /{ (7/4){(4/7)[x - (1/2)]² + 1} } =

(1/4) ln |x + 1| - (1/8) ln (x² - x + 2) + (3/8)(4/7) ∫ dx /{(4/7)[x - (1/2)]² + 1} =

include (4/7) into the square as (2/√7):

(1/4) ln |x + 1| - (1/8) ln (x² - x + 2) + (3/14) ∫ dx /{ {(2/√7)[x - (1/2)]}² + 1} } =

expand the base of the square as:

(1/4) ln |x + 1| - (1/8) ln (x² - x + 2) + (3/14) ∫ dx / {[(2/√7)x - (1/√7)]² + 1} =

finally, divide and multiply the remaining integral by (2/√7), in order to change the

numerator into the derivative of [(2/√7)x - (1/√7)] (that is (2/√7)):

(1/4) ln |x + 1| - (1/8) ln (x² - x + 2) +

(3/14)[(√7)/2] ∫ (2/√7)dx / {[(2/√7)x - (1/√7)]² + 1} =

(1/4) ln |x + 1| - (1/8) ln (x²- x + 2) +

[3/(4√7)] ∫ {d[(2/√7)x - (1/√7)]} / {[(2/√7)x - (1/√7)]² + 1} =

note that the last integral is of the type: ∫ {d[f(x)]} /{[f(x)]² + 1} = arctan[f(x)] + C

therefore, in conclusion, you get:

∫ [1 / (x³ + x + 2)] dx =

(1/4) ln|x + 1| - (1/8) ln(x²- x + 2) + [3/(4√7)] arctan[(2/√7)x - (1/√7)] + C

very interesting integral...

I hope it helps....

Bye!

Answer 2

x^3 + x + 2

= x^3 + 1 + x + 1

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

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

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

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

x = -1 => A = 1/4

Comparing coefficients of x^2 and constant terms,

A + B = 0 => B = -A = - 1/4 and

2A + C = 1 => C = 1 - 2A = 1/2

=> ∫1/(x^3+x+2) dx

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

= (1/4) ln lx + 1l - (1/8)∫(2x - 1 - 3) / (x^2 - x+ 2) dx

= (1/4) ln lx + 1l - (1/8)∫(2x - 1) / (x^2 - x+ 2) dx -(3/8)∫dx / [(x -1/2)^2 - (√7/2)^2]

= (1/4) ln lx + 1l - (1/8)ln lx^2 - x + 2l - (3/4√7) ln l (2x - 1 - √7) / (2x - 1 + √7) + c

(Some typo errors have been edited)

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