If ∫3x+1(x−1)3(x+1)dx=A.logx+1x−1+Bx−1+C(x−1)2+D then A+B+C=

∫3x+1(x-1)3 (x+1)dx 3x+1(x-1)3 (x+1)=Ax-1+Bx+12+C(x-1)3+Dx+1 3x+1=A(x-1)2-(x+1)+B(x-1)(x+1)+C(x+1)+D(x-1)3 x=1 ⇒ 4=0+0+2C+0 C=2 x=-1 ⇒ -2=0+0+0+0(-2)3 D=+1/4 x2 coef ⇒ 0=-A+B+d(-3) 34=-A+B x2 coef ⇒ 0=A+D A=1/4 B=1/2 I=∫-1/4x-1+1/2(x-1)2+2(x-1)3+1/4x+1 =-14log |x-1|+-12(x-1)+-1(x-1)2+14log |x+1| =14 log x+1x-1- 12(x-1)-1(x-1)2 A+B+C=14-12-1=-5/4

If ∫3x+1(x−1)3(x+1)dx=A.logx+1x−1+Bx−1+C(x−1)2+D then A+B+C=

∫3x+1(x-1)3 (x+1)dx 3x+1(x-1)3 (x+1)=Ax-1+Bx+12+C(x-1)3+Dx+1 3x+1=A(x-1)2-(x+1)+B(x-1)(x+1)+C(x+1)+D(x-1)3 x=1 ⇒ 4=0+0+2C+0 C=2 x=-1 ⇒ -2=0+0+0+0(-2)3 D=+1/4 x2 coef ⇒ 0=-A+B+d(-3) 34=-A+B x2 coef ⇒ 0=A+D A=1/4 B=1/2 I=∫-1/4x-1+1/2(x-1)2+2(x-1)3+1/4x+1 =-14log |x-1|+-12(x-1)+-1(x-1)2+14log |x+1| =14 log x+1x-1- 12(x-1)-1(x-1)2 A+B+C=14-12-1=-5/4

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If $\int \frac{3 x + 1}{{(x - 1)}^{3} (x + 1)} d x = A . \log |\frac{x + 1}{x - 1}| + \frac{B}{x - 1} + \frac{C}{{(x - 1)}^{2}} + D t h e n A + B + C =$

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$\frac{5}{2}$

$\frac{5}{4}$

$\frac{- 5}{4}$

$\frac{- 5}{2}$

answer is A.

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Detailed Solution

$\int \frac{3 x + 1}{{(x - 1)}^{3} (x + 1)} dx \frac{3 x + 1}{{(x - 1)}^{3} (x + 1)} = \frac{A}{x - 1} + \frac{B}{{(x + 1)}^{2}} + \frac{C}{{(x - 1)}^{3}} + \frac{D}{x + 1} 3 x + 1 = A {(x - 1)}^{2} - (x + 1) + B (x - 1) (x + 1) + C (x + 1) + D {(x - 1)}^{3} x = 1 \Rightarrow 4 = 0 + 0 + 2 C + 0 C = 2 x = - 1 \Rightarrow - 2 = 0 + 0 + 0 + 0 {(- 2)}^{3} D = + 1 / 4 x^{2} coef \Rightarrow 0 = - A + B + d (- 3) \frac{3}{4} = - A + B x^{2} coef \Rightarrow 0 = A + D A = 1 / 4 B = 1 / 2 I = \int (\frac{- 1 / 4}{x - 1} + \frac{1 / 2}{{(x - 1)}^{2}} + \frac{2}{{(x - 1)}^{3}} + \frac{1 / 4}{x + 1}) = \frac{- 1}{4} \log | x - 1 | + \frac{- 1}{2 (x - 1)} + \frac{- 1}{{(x - 1)}^{2}} + \frac{1}{4} \log | x + 1 | = \frac{1}{4} \log |\frac{x + 1}{x - 1}| - \frac{1}{2 (x - 1)} - \frac{1}{{(x - 1)}^{2}} A + B + C = \frac{1}{4} - \frac{1}{2} - 1 = - 5 / 4$