Answer :

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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