The value of ^3 x ( ^4 x + 3 ^2 x + 1) ⁻¹( x + x) ,dx is

Step-by-step solution

Let t = x + x . Differentiating with respect to x , we get: dt = ( x x - x) dx dt = x ( 1 ^2 x - 1 ) dx = ^3 x ^2 x dx The polynomial in the denominator can be rewritten as: ^4 x + 3 ^2 x + 1 = ^2 x ( ^2 x + 3 + ^2 x) ^4 x + 3 ^2 x + 1 = ^2 x (( x + x)^2 + 1) = ^2 x (t^2 + 1) Substituting these into the given integral, we get: I = ^3 x ^2 x (t^2 + 1) ⁻¹ t dx I = dt (t^2 + 1) ⁻¹ t Now, let u = ⁻¹ t . Differentiating with respect to t , we get: du = dt t^2 + 1 The integral simplifies to: I = du u = (u) + c Substituti