How do you find the indefinite integral of 1/(x^2+x+1)^2 ?

Question

How do you find the indefinite integral of 1/(x^2+x+1)^2 ?

Answer 1

Its a very long process so I shall give you steps.

x^2 + x + 1 = [ x+(1/2)]^2 +[(rt3)/2]^2

so substitute.x+(1/2) = [(rt3)/2]tanU,

YOU will get integral of (2/3)dU/[secU]^4= (2/3)[cosU]^4dU

Now use cos2T = 2(c0sT)^2 - 1 twice to get [cosU]^4 in terms of cosines of multiples of U, then integrate and re-substitute.

Answer 2

integral of 1/(x^2+x+1)^2 with respect to x is

=> ((3 + 6*x)/(1 + x + x^2) + (2*i)*sqrt(3)*log(-i + sqrt(3) - (2*i)*x) - (2*i)*sqrt(3)*log(i + sqrt(3) + (2*i)*x))/9 + Costant

where i = sqrt(-1), an imaginary number (iota).

Answer 3

very tough integral...

∫ [1 /(x² + x + 1)²] dx =

note that the base of the square is unfactorable; thus you have to attempt to

complete the square:

x² + x + 1 =

split 1 into (1/4) + (3/4):

[x² + x + (1/4)] + (3/4) =

[x + (1/2)]² + (3/4)

the integrand thus becoming:

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

divide and multiply by (3/4), yielding:

(4/3) ∫ { (3/4) /{ [x + (1/2)]² + (3/4) }² } dx =

add {[x + (1/2)]² - [x + (1/2)]²} to the numerator:

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

then distribute and split as:

(4/3) ∫ { { {(3/4) + [x + (1/2)]²} /{ [x + (1/2)]² + (3/4) }²} -

{ [x + (1/2)]² /{ [x + (1/2)]² + (3/4) }²} } dx =

(4/3) ∫ { {(3/4) + [x + (1/2)]²} /{ [x + (1/2)]² + (3/4) }²} dx -

(4/3) ∫ { [x + (1/2)]² /{ [x + (1/2)]² + (3/4) }²} dx =

the first integrand simplifies into:

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

rewrite the second integral as:

(4/3) ∫ {1 /{ [x + (1/2)]² + (3/4) }} dx -

(4/3) ∫ [x + (1/2)] { [x + (1/2)] /{ [x + (1/2)]² + (3/4) }²} dx =

then integrate it by parts, assuming:

[x + (1/2)] = u → dx = du

{ [x + (1/2)] /{ [x + (1/2)]² + (3/4) }²} dx = dv →

(1/2) (2) { [x + (1/2)] /{ [x + (1/2)]² + (3/4) }²} dx = dv →

(1/2) { {2[x + (1/2)] dx} /{ [x + (1/2)]² + (3/4) }²} = dv →

note that 2[x + (1/2)] dx = d { [x + (1/2)]² + (3/4) } →

(1/2) { d { [x + (1/2)]² + (3/4) }} /{ [x + (1/2)]² + (3/4) }² = dv →

(1/2) {{ [x + (1/2)]² + (3/4) }^(-2)} d { [x + (1/2)]² + (3/4) } = dv →

integrating,

(1/2) {{ [x + (1/2)]² + (3/4) }^(-2+1)} /(-2+1) = v →

(1/2) {{ [x + (1/2)]² + (3/4) }^(-1)} /(-1) = v →

(-1/2) {1/{ [x + (1/2)]² + (3/4) } } = v

thus, solving the second integral by parts (see above), you get:

(4/3) ∫ {1 /{ [x + (1/2)]² + (3/4) }} dx -

(4/3) ∫ [x + (1/2)] { [x + (1/2)] /{ [x + (1/2)]² + (3/4) }²} dx =

(4/3) ∫ {1 /{ [x + (1/2)]² + (3/4) }} dx - (4/3) { (-1/2) {1/{ [x + (1/2)]² + (3/4) }[x + (1/2)] } -

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

(4/3) ∫ {1 /{ [x + (1/2)]² + (3/4) }} dx - (4/3) { (-1/2) {[x + (1/2)]/{ [x + (1/2)]² + (3/4) } } +

(1/2) ∫ {1/{ [x + (1/2)]² + (3/4) } } dx =

(4/3) ∫ {1 /{ [x + (1/2)]² + (3/4) }} dx - (4/3)(-1/2) {[x + (1/2)]/{ [x + (1/2)]² + (3/4) } } -

(4/3)(1/2) ∫ {1/{ [x + (1/2)]² + (3/4) } } dx =

(4/3) ∫ {1 /{ [x + (1/2)]² + (3/4) }} dx + (2/3){[x + (1/2)]/ { [x + (1/2)]² + (3/4) } } -

(2/3) ∫ {1/{ [x + (1/2)]² + (3/4) }} dx =

note that the remaining integrals are similar, thus:

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

(2/3) ∫ {1 /{ [x + (1/2)]² + (3/4) }} dx + (2/3){[x + (1/2)]/ { [x + (1/2)]² + (3/4) }} =

as for the remaining integral, factor out 3/4 from the denominator:

(2/3) ∫ dx /{(3/4) {(4/3)[x + (1/2)]² + 1}} + (2/3){[x + (1/2)]/ { [x + (1/2)]² + (3/4) }} =

taking (3/4) out:

(2/3)(4/3) ∫ dx /{(4/3)[x + (1/2)]² + 1} + (2/3){[x + (1/2)]/ { [x + (1/2)]² + (3/4) }} =

then include (4/3) in the square as:

(8/9) ∫ dx /{ {(2/√3)[x + (1/2)]}² + 1} + (2/3){[x + (1/2)]/ { [x + (1/2)]² + (3/4) }} =

expand the base of the square as:

(8/9) ∫ dx /{ {[(2/√3)x] + [(2/√3)(1/2)]}² + 1} + (2/3){[x + (1/2)]/ { [x + (1/2)]² + (3/4) }} =

(8/9) ∫ dx /{[(2/√3)x + (1/√3)]² + 1} + (2/3){[x + (1/2)]/ { [x + (1/2)]² + (3/4) }} =

finally, in order to turn the numerator of the integrand into the derivative

of [(2/√3)x + (1/√3)] (i.e. 2/√3), divide and multiply by (2/√3), yielding:

(8/9)[(√3)/2] ∫ (2/√3)dx /{[(2/√3)x + (1/√3)]² + 1} +

(2/3){[x + (1/2)]/ { [x + (1/2)]² + (3/4) }} =

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

(2/3){[x + (1/2)]/ { [x + (1/2)]² + (3/4)}} =

note that the remaining integral has been rearranged into the form

∫ {d [f(x)]} /{[f(x)]²+ 1} = arctan[f(x)] + C, therefore:

[4/(3√3)] arctan[(2/√3)x + (1/√3)] + (2/3){[x + (1/2)]/ { [x + (1/2)]² + (3/4) }} + C =

[4/(3√3)] arctan[(2/√3)x + (1/√3)] + (2/3){[x + (1/2)]/ { [x² + x + (1/4)] + (3/4) }} + C =

[4/(3√3)] arctan[(2/√3)x + (1/√3)] + (2/3){[x + (1/2)]/ (x² + x + 1)} + C

in conclusion,

∫ [1 /(x²+ x +1)²] dx = [4/(3√3)]arctan[(2/√3)x + (1/√3)] + (2/3){[x + (1/2)]/(x²+ x +1)} + C

somewhat tough.....nevertheless interesting

I hope this helps...

Bye!