How do you evaluate the integral $int (3x+6)/(x^4-2x^2+1)$ ?

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Answer 1 · 2018-03-24T17:15:44

$3/2Ln((x+1)/(x-1))-(6x+3)/(2x^2-2)+C$

$int (3x+6)/(x^4-2x^2+1)*dx$

= $int (3x+6)/(x^2-1)^2*dx$

= $int (3x+6)/[(x+1)*(x-1)]^2*dx$

Now I decomposed integrand into basic fractions,

$(3x+6)/[(x+1)*(x-1)]^2=A/(x+1)+B/(x+1)^2+C/(x-1)+D/(x-1)^2$

After expanding denominator,

$A(x+1)(x-1)^2+B(x-1)^2+C(x-1)(x+1)^2+D(x+1)^2=3x+6$ or

$A*(x^3-x^2-x+1)+B*(x^2-2x+1)+C*(x^3+x^2-x-1)+D*(x^2+2x+1)=3x+6$

$(A+C)*x^3+(-A+B+C+D)*x^2+(-A-2B-C+2D)*x+A+B-C+D=3x+6$

After equating coefficients, I found

$A+C=0$

$-A+B+C+D=0$

$-A-2B-C+2D=3$ and

$A+B-C+D=6$

After solving them, I found $A=3/2$ , $C=-3/2$ , $D=9/4$ and $B=3/4$

Thus,

$int (3x+6)/[(x+1)*(x-1)]^2*dx$

= $3/2int (dx)/(x+1)+3/4int (dx)/(x+1)^2-3/2int (dx)/(x-1)+9/4int (dx)/(x-1)^2$

= $3/2Ln(x+1)-3/(4x+4)-3/2Ln(x-1)-9/(4x-4)+C$

= $3/2Ln((x+1)/(x-1))-(6x+3)/(2x^2-2)+C$

How do you evaluate the integral int (3x+6)/(x^4-2x^2+1)int (3x+6)/(x^4-2x^2+1)?