Q.

If (1x+1x3)(3x24+x2623)dx=α3(α+1)(3xβ+xγ)α+1α+C, x > 0, (α,β,γZ), where c is the constant of integration, then α+β+γ is equal to __________.          [2025]


Ans.

(19)

We have, (1x2+1x4)(3x+1x3)123dx

Put t=3x+1x3  dt=3(1x2+1x4)dx

Now, t1/23dt3=t24/23(2423)(3)+C

On comparing, we get α=23, β=1, γ=3

  α+β+γ=19.