Find:

Let I =

Let

⇒ (2x + 1) = Ax(x² + 4) + B(x² + 4) + Cx(x² + 1) + D(x² + 1)

⇒ (2x + 1) = Ax³ + 4Ax + Bx² + 4B + Cx³ + Cx + Dx² + D

⇒ 2x + 1 = x³(A + C) + x²(B + D) + x(4A +C) + 1(4B + D)

On comparing the coefficient of x and 1, we get

A + C = 0 …(i)

B + D = 0 …(ii)

4A + C = 2 …(iii)

4B + D = 1 …(iv)

Solving eq. (i) and (iii), we get

4A + C – A – C = 2 – 0

⇒ 3A = 2

Now, solving eq. (ii) and (iv), we get

4B + D – B – D = 1 – 0

⇒ 3B = 1

Put in eq. (i), we get

Put in eq. (ii), we get

So, we get

Solving I₁,

Let x² + 1 = t

On differentiating, we get

⇒ 2x dx = dt

Solving I₂,

Now, we know that,

So,

Solving I₃,

Let x² + 4 = t

On differentiating, we get

⇒ 2x dx = dt

Solving I₄,

Now, we know that,

So,

Putting the value of I₁, I₂, I₃, I₄ in eq. (i), we get

∵C₁, C₂, C₃, C₄ are constants.

Let C₁ + C₂ + C₃ + C₄ = C

∴