If ∫4(3−2x)−23−2x3+2x13dx=3α3+2x3−2xβ/γ+C (β&γ are prime numbers). then

I=∫4(3−2x)23-2x3+2x13dxPutting 3+2x3−2x=t ⇒123−2x2dx=dt ⇒I=13∫t-13dt =12t23+c =123+2x3−2x23+c α=6,β=2,γ=3

If ∫4(3−2x)−23−2x3+2x13dx=3α3+2x3−2xβ/γ+C (β&γ are prime numbers). then

I=∫4(3−2x)23-2x3+2x13dxPutting 3+2x3−2x=t ⇒123−2x2dx=dt ⇒I=13∫t-13dt =12t23+c =123+2x3−2x23+c α=6,β=2,γ=3

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If $\int 4 (3 - 2 x)^{- 2} {(\frac{3 - 2 x}{3 + 2 x})}^{\frac{1}{3}} dx = \frac{3}{α} {(\frac{3 + 2 x}{3 - 2 x})}^{β / γ} + C$ ( $β & γ$ are prime numbers). then

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$α, β, γ are in G.P$

$α, β, γ are in H.P$

$α, 2 β, γ are in H.P$

$α = βγ$

answer is C, D.

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Detailed Solution

$I = \int \frac{4}{(3 - 2 x)^{2}} {(\frac{3 - 2 x}{3 + 2 x})}^{\frac{1}{3}} dx$

Putting
$\frac{3 + 2 x}{3 - 2 x} = t \Rightarrow \frac{12}{{(3 - 2 x)}^{2}} dx = d t \Rightarrow I = \frac{1}{3} \int t^{\frac{- 1}{3}} d t = \frac{1}{2} t^{\frac{2}{3}} + c = \frac{1}{2} {(\frac{3 + 2 x}{3 - 2 x})}^{\frac{2}{3}} + c α = 6, β = 2, γ = 3$