If ∫(2x+3)x(x+1)(x+2)(x+3)+1dx=−1px2+qx+r+c then 3p-qr=

∫ 2x+3x(x+1) (x+2) (x+3)+1dx ∫ 2x+3(x2+3x+2) (x2+3x)+1 Put x2+3x=t =∫ dt(t+2) t+1 =∫ dtt2+2t+1=∫ dt(t+1)2 =-1t+1+c =-1x2+3x+1+c 3p-qr=3(1)-31=0

If ∫(2x+3)x(x+1)(x+2)(x+3)+1dx=−1px2+qx+r+c then 3p-qr=

∫ 2x+3x(x+1) (x+2) (x+3)+1dx ∫ 2x+3(x2+3x+2) (x2+3x)+1 Put x2+3x=t =∫ dt(t+2) t+1 =∫ dtt2+2t+1=∫ dt(t+1)2 =-1t+1+c =-1x2+3x+1+c 3p-qr=3(1)-31=0

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If $\int \frac{(2 x + 3)}{x (x + 1) (x + 2) (x + 3) + 1} d x = - \frac{1}{p x^{2} + q x + r} + c t h e n \frac{3 p - q}{r} =$

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Detailed Solution

$\int \frac{2 x + 3}{x (x + 1) (x + 2) (x + 3) + 1} dx \int \frac{2 x + 3}{(x^{2} + 3 x + 2) (x^{2} + 3 x) + 1} Put x^{2} + 3 x = t = \int \frac{dt}{(t + 2) t + 1} = \int \frac{dt}{t^{2} + 2 t + 1} = \int \frac{dt}{{(t + 1)}^{2}} = \frac{- 1}{t + 1} + c = - \frac{1}{x^{2} + 3 x + 1} + c \frac{3 p - q}{r} = \frac{3 (1) - 3}{1} = 0$