integral x22x1x12x21Solution x 2x 1 x 1x 1 dx since the

Solution

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

since the denominator is completely factored you can start with partial
fraction decomposition as:

(x² - 2x - 1) /[(x - 1)²(x²+ 1)] = A/(x - 1) + B/(x - 1)² + (Cx + D)/(x²+ 1)

both (x - 1) and (x - 1)² denominators are needed because (x - 1) is squared
let [(x - 1)²(x²+ 1)] be the common denominator:

(x² - 2x - 1) /[(x - 1)²(x²+ 1)] =
[A(x - 1)(x²+ 1) + B(x²+ 1) + (Cx + D)(x - 1)²] /[(x - 1)²(x²+ 1)]

equate the numerators:

(x² - 2x - 1) = [A(x - 1)(x²+ 1) + B(x²+ 1) + (Cx + D)(x - 1)²]

x² - 2x - 1 = A(x³+ x - x²- 1) + Bx²+ B + (Cx + D)(x² - 2x + 1)

x² - 2x - 1 = Ax³+ Ax - Ax²- A + Bx²+ B + Cx³ - 2Cx² + Cx + Dx² - 2Dx + D
x² - 2x - 1 = (A + C)x³+ (- A + B - 2C + D)x² + (A + C - 2D)x + (- A + B + D)
yielding the four unknowns system:

| A + C = 0
| - A + B - 2C + D = 1
| A + C - 2D = - 2
| - A + B + D = - 1

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

| A = - C = 1
| D = 1
| - C + B + 1 = 1 B = 1 - 1 + C B = C = - 1
| C + C + 1 = - 1 2C = - 2 C = - 2/2 = - 1

thus: A = 1, B = - 1, C = - 1, D = 1 so that (see above):

(x² - 2x - 1) /[(x - 1)²(x²+ 1)] = A/(x - 1) + B/(x - 1)² + (Cx + D)/(x²+ 1)

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

the given integral thus becoming:

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

breaking it up,

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

breaking it up furtherly,

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

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

rewrite the second integral as a negative power and divide and multiply
the third integral by 2, so that the numerator turns into the derivative of the denominator:

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

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

integrating all terms:

ln |x - 1| - [(x - 1)^(-2+1)]/(-2+1) - (1/2) ln (x²+ 1) + arctan x + C =

ln |x - 1| - [(x - 1)^(-1)]/(-1) - (1/2) ln (x²+ 1) + arctan x + C

thus, in conclusion:

{(x²- 2x -1) /[(x -1)²(x²+1)]} dx = ln |x - 1| + [1/ (x -1)] - (1/2) ln (x²+1) + arctan x + C