Sol. Let I = dx

Let x + = t, then, = t –x

⇒ x 2 –x + 1 = (t –x) 2 ⇒ – x + 1 = t

2 –2tx

⇒ x =

∴ dx = dt = dt

Substituting these values, we get

I = dt = 2 dt

Let = + + ... (i)

Multiplying both sides by t and putting t = 0, we get A = 1

Similarly, multiplying both sides of (i) by

(2t –1) 2 and putting t = , we get

C = =

From (i), we have

t 2 –t + 1 = A (2t –1) 2 + B (2t –1) t + Ct

On equating coefficient of t 2 on both sides, we get

1 = 4A + 2B ⇒ B = –

Substituting the values of A, B, C in (i), we get

= – +

∴ I = 2 dt –3 dt + dt

⇒ I = 2 log t – log (2t –1) – + C

⇒ I = 2 log (x + ) – log

– + C