Evaluate the integral (x^2+3)/(x^2+2x+3)^2 using partial fractions?

Question

Evaluate the integral (x^2+3)/(x^2+2x+3)^2 using partial fractions?

Answer 1

very tough integral.....

∫ [(x²+ 3) / (x²+ 2x +3)²] dx =

the denominator is already completely factored, thus you can set

partial fraction decomposition as:

(x²+ 3)/(x²+ 2x +3)² = (Ax + B)/(x²+ 2x +3) + (Cx + D)/(x²+ 2x +3)² →

find out the LCD at right side:

(x²+ 3) /(x²+ 2x +3)² = [(Ax + B)(x²+ 2x +3) + (Cx + D)] /(x²+ 2x +3)² →

equate the numerators:

(x²+ 3) = [(Ax + B)(x²+ 2x +3) + (Cx + D)] →

x² + 3 = Ax³ + 2Ax² + 3Ax+ Bx²+ 2Bx +3B + Cx + D →

x² + 3 = Ax³ + (2A + B)x²+ (3A + 2B + C)x + (3B + D)

yielding the four unknowns system:

| A = 0

| 2A + B = 1

| 3A + 2B + C = 0

| 3B + D = 3

| A = 0

| 0 + B = 1 → B = 1

| 0 + 2(1) + C = 0 → C = - 2

| 3B + D = 3 → D = 3 - 3(1) = 0

| A = 0

| B = 1

| C = - 2

| D = 0

therefore (see above):

(x²+ 3)/(x²+ 2x +3)² = (Ax + B)/(x²+ 2x +3) + (Cx + D)/(x²+ 2x +3)² →

(x²+ 3)/(x²+ 2x +3)² =  1 /(x²+ 2x +3) - 2x /(x²+ 2x +3)²

your integral thus becoming:

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

break it up into:

∫ [1 /(x² + 2x + 3)] dx - ∫ [2x /(x²+ 2x +3)²] dx =

now, attempting to turn the numerator of the second integrand into the derivative

of the square base in the respective denominator, add (2 - 2) to the numerator:

∫ [1 /(x² + 2x + 3)] dx - ∫ [(2x + 2 - 2)/(x²+ 2x +3)²] dx =

split it into:

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

∫ [1 /(x² + 2x + 3)] dx - ∫ [(2x + 2)/(x²+ 2x +3)²] dx + ∫ [2 /(x²+ 2x +3)²] dx =

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

∫ [1 /(x² + 2x + 3)] dx - ∫ (x²+ 2x +3)^(-2) d(x²+ 2x +3) + ∫ [2 /(x²+ 2x +3)²] dx =

∫ [1 /(x² + 2x + 3)] dx - [(x²+ 2x +3)^(-2+1)]/(-2+1) + ∫ [2 /(x²+ 2x +3)²] dx =

∫ [1 /(x² + 2x + 3)] dx - [(x²+ 2x +3)^(-1)]/(-1) + ∫ [2 /(x²+ 2x +3)²] dx =

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

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

let's now complete the square in the denominator of the second integral:

x²+ 2x + 3 =

(x²+ 2x +1) + 2 =

(x + 1)² + 2

thus:

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

add [(x + 1)²- (x + 1)²] to the numerator of the second integrand:

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

then distribute and split as:

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

[1 /(x²+ 2x + 3)] =

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

[1 /(x²+ 2x + 3)] =

simplify the second integrand:

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

[1 /(x²+ 2x + 3)] =

then expand it into:

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

[1 /(x²+ 2x + 3)] =

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

[1 /(x²+ 2x + 3)] =

the first and the second integral are similar, thus add them together into:

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

[1 /(x²+ 2x + 3)] =

rewrite the denominator of the first integral as (see above):

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

[1 /(x²+ 2x + 3)] =

then factor out 2 from the denominator:

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

[1 /(x²+ 2x + 3)] =

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

[1 /(x²+ 2x + 3)] =

include (1/2) in the square as:

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

[1 /(x²+ 2x + 3)] =

expand the base of the square into:

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

[1 /(x²+ 2x + 3)] =

then multiply and divide the first integral by √2, in order to turn the numerator

into the derivative of [(x/√2) + (1/√2)] (i.e. (1/√2)):

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

[1 /(x²+ 2x + 3)] =

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

[1 /(x²+ 2x + 3)] =

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

as for the remaining integral, rewrite it as:

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

and integrate it by parts, assuming:

(x + 1) = u → dx = du

{(x + 1) /[(x + 1)²+ 2]²} dx = dv →

(divide and multiply by 2)

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

note that 2(x + 1) dx is the same as d[(x + 1)²+ 2], thus

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

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

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

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

(-1/2) {1/ [(x + 1)²+ 2]} = v

then, integrating the remaining integral by parts, you get:

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

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

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

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

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

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

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

similarly, factor out 2 from the denominator of the remaining integral:

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

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

then include (1/2) in the square as (1√2):

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

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

multiply and divide the remaining integral by √2:

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

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

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

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

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

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

expanding arctan argument:

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

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

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

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

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

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

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

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

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

thus, in conclusion:

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

very tough.....but interesting....

I hope it helps....

Bye!