How do you integrate $int 2/((x^4-1)dx$ using partial fractions?

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Answer 1 · 2016-05-05T05:39:57

$int 2/(x^4-1) dx=-tan^-1 x-1/2ln(x+1)+1/2ln(x-1)+C_0$

$int 2/(x^4-1) dx$

for the partial fraction

$2/(x^4-1) =2/((x^2+1)(x^2-1))=(Ax+B)/(x^2+1)+C/(x+1)+D/(x-1)$

$2=(Ax+B)(x^2-1)+C(x^2+1)(x-1)+D(x^2+1)(x+1)$

We now have the equation

$2=Ax^3+Bx^2-Ax-B+C(x^3+x-x^2-1)+D(x^3+x+x^2+1)$

and then

$A+C+D=0" "$ first equation
$B-C+D=0" "$ second equation
$-A+C+D=0" "$ third equation
$-B-C+D=2" "$ fourth equation

simultaneous solution results to

$A=0$
$B=-1$
$C=-1/2$
$D=1/2$

$int 2/(x^4-1) dx =int(Ax+B)/(x^2+1) dx+int C/(x+1) dx+int D/(x-1) dx$

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

$int 2/(x^4-1) dx=-tan^-1 x-1/2ln(x+1)+1/2ln(x-1)+C_0$

God bless....I hope the explanation is useful.

How do you integrate int 2/((x^4-1)dxint 2/((x^4-1)dx using partial fractions?