Evaluate the foll

Denominator is factorized, so let separate the fraction through partial fraction, hence let

⇒ x = (Ax + B)(x – 1) + (Cx + D)(x² + 1)

⇒ x = Ax² – Ax + Bx – B + Cx² + Cx + Dx² + D

⇒ x = (C) x² + (A + D) x² + (B – A + C)x + (D – B)……(ii)

By equating similar terms, we get

C = 0 ………..(iii)

A + D = 0⇒ A = – D …………(iv)

B – A + C = 1

⇒ B – ( – D) + 0 = 2 (from equation(iii) and (iv))

⇒ B = 2 – D………..(v)

D – B = 0 ⇒ D – (2 – D) = 0 ⇒ 2D = 2 ⇒ D = 1

So equation(iv) becomes A = – 1

So equation (v) becomes, B = 2 – 1 = 1

We put the values of A, B, C, and D values back into our partial fractions in equation (i) and replace this as the integrand. We get

Split up the integral,

Let substitute

so the above equation becomes,

On integrating we get

(the standard integral of )

Substituting back, we get

Note: the absolute value signs account for the domain of the natural log function (x>0).

Hence,