Question

Evaluate:

`int x^2/((x - 1)^2(x^2 + 1)) dx`

Answer 1

Let I = `int x^2/((x - 1)^2(x^2 + 1)) dx`

By partial fraction,

`x^2/((x - 1)^2(x^2 + 1)) = A/((x - 1)) + B/(x - 1)^2 + (Cx + D)/((x^2 + 1))`

x² = A(x – 1) (x² + 1) + B(x² + 1) + (Cx + D) (x – 1)²   ...(i)

Put x – 1 = 0

⇒ x = 1

1 = B(2)

⇒ `B = 1/2`

Compare the coefficient of x³ in equation (i),

0 = A + C   ...(ii)

Compare the coefficient of x² in equation (i),

1 = – A + B – 2C + D   ...(iii)

Compare the coefficient of x in equation (i),

0 = A + C – 2D   ...(iv)

Put the value of (A + C) in equation (iv)

0 = 0 – 2D

∴ D = 0

Compare the coefficient of constant term

0 = – A + B + D

`0 = -A + 1/2 + 0`

⇒ `A = 1/2`

And `C = -1/2`

Now `int x^2/((x - 1)^2(x^2 + 1)) dx`

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

`I = 1/2 log |x - 1| + 1/2 (x - 1)^-1/(-1) - 1/2 int x/(x^2 + 1) dx`

Put x² + 1 = t

2x dx = dt

`x  dx = 1/2 dt`

`I = 1/2 log |x - 1| - 1/(2(x - 1)) - 1/4 int dt/t + C`

`I = 1/2 log |x - 1| - 1/(2(x - 1)) - 1/4 log |x^2 + 1| + C`