Answer / nirali

3x+5/x^3-x^2-x+1
--->3x+5/x^2(x-1)-(x-1)
=3x+5/(x-1)(x^2-1)
=3x+5/(x-1)^2(x+1)
Let 3x+5/(x-1)^2(x+1)=A/(x-1) +B/(x-1)^2 +C/x+1
Then,
3x+5=A(x-1)(x+1)+B(x+1)+C(x-1)^2......(i)
Putting x=1,-1 and 0 successively in (i),we get
8=2B,2=4C and 5= -A+B+C
--->B=4,C=1/2,A=-1/2
Therefore,
3x+5/(x-1)^2(x+1)=-1/2(x-1) +4/(x-1)^2 +1/2(x+1)
--->integration of 3x+5/(x-1)^2(x+1) =-1/2 log|x-1| -4/x-1 +1/2log|x+1|+ c
=1/2log|x+1/x-1| -4/x-1 +c

Answer / sivka

3x+5/x^3-x^2-x+1
--->3x+5/x^2(x-1)-(x-1)
=3x+5/(x-1)(x^2-1)
=3x+5/(x-1)^2(x+1)
Let 3x+5/(x-1)^2(x+1)=A/(x-1) +B/(x-1)^2 +C/x+1
Then,
3x+5=A(x-1)(x+1)+B(x+1)+C(x-1)^2......(i)
Putting x=1,-1 and 0 successively in (i),we get
8=2B,2=4C and 5= -A+B+C
--->B=4,C=1/2,A=-1/2
Therefore,
3x+5/(x-1)^2(x+1)=-1/2(x-1) +4/(x-1)^2 +1/2(x+1)
--->integration of 3x+5/(x-1)^2(x+1) =-1/2 log|x-1| -4/x-1 +1/2log|x+1|+ c
=1/2log|x+1/x-1| -4/x-1 +c

Answer / ankit

3/(x-1)^2 + 2/ {(x-1)^2*(x+1)}
-3/(x-1) - 2*t*dt/(1-2t)^2 if x=1/t -1
-3/(x-1) + dt/1-2t - dt /(1-2t)^2
-3/(x-1) -1/2log(1-2t) - 1/{2*(1-2t)}
-3/x-1 -1/2 log (1-2/x+1) - 1/2(1-2/x+1)
-3/(x-1) -1/2 log{(x-1)/(x+1)} -(x+1)/{2*(x-1)}

-(x+7)/{2*(x-1)} -1/2 log{(x-1)/(x+1)}

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