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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

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