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

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

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Answer 1 · 2016-05-05T05:39:57

$\int \frac{2}{x^{4} - 1} d x = - {tan}^{- 1} x - \frac{1}{2} ln (x + 1) + \frac{1}{2} ln (x - 1) + C_{0}$ $int 2/(x^4-1) dx=-tan^-1 x-1/2ln(x+1)+1/2ln(x-1)+C_0$

Explanation:

$\int \frac{2}{x^{4} - 1} d x$ $int 2/(x^4-1) dx$

for the partial fraction

$\frac{2}{x^{4} - 1} = \frac{2}{(x^{2} + 1) (x^{2} - 1)} = \frac{A x + B}{x^{2} + 1} + \frac{C}{x + 1} + \frac{D}{x - 1}$ $2/(x^4-1) =2/((x^2+1)(x^2-1))=(Ax+B)/(x^2+1)+C/(x+1)+D/(x-1)$

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

We now have the equation

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

and then

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

simultaneous solution results to

$A = 0$ $A=0$
$B = - 1$ $B=-1$
$C = - \frac{1}{2}$ $C=-1/2$
$D = \frac{1}{2}$ $D=1/2$

$\int \frac{2}{x^{4} - 1} d x = \int \frac{A x + B}{x^{2} + 1} d x + \int \frac{C}{x + 1} d x + \int \frac{D}{x - 1} d x$ $int 2/(x^4-1) dx =int(Ax+B)/(x^2+1) dx+int C/(x+1) dx+int D/(x-1) dx$

$\int \frac{2}{x^{4} - 1} d x = \int \frac{- 1}{x^{2} + 1} d x + \int \frac{- \frac{1}{2}}{x + 1} d x + \int \frac{\frac{1}{2}}{x - 1} d x$ $int 2/(x^4-1) dx =int(-1)/(x^2+1) dx+int (-1/2)/(x+1) dx+int (1/2)/(x-1) dx$

$\int \frac{2}{x^{4} - 1} d x = - {tan}^{- 1} x - \frac{1}{2} ln (x + 1) + \frac{1}{2} ln (x - 1) + C_{0}$ $int 2/(x^4-1) dx=-tan^-1 x-1/2ln(x+1)+1/2ln(x-1)+C_0$

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

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

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