We have an answer to "Evaluate the following integrals. ∫dxx4−1"

bobbie71G

Answered question

2021-10-16

Evaluate the following integrals.

\int \frac{d x}{x^{4} - 1}

Answer & Explanation

Nathanael Webber

Skilled2021-10-17Added 117 answers

Evaluate:
$\int \frac{d x}{x^{4} - 1}$
Simplification:
$\int \frac{d x}{x^{4} - 1} = \int \frac{d x}{(x^{2} + 1) (x^{2} - 1)}$
$= \int \frac{d x}{(x^{2} + 1) (x + 1) (x - 1)}$
Now simplify $\frac{1}{(x^{2} + 1) (x + 1) (x - 1)}$ using partial fraction method,
$\frac{1}{(x^{2} + 1) (x + 1) (x - 1)} = \frac{A x + B}{x^{2} + 1} + \frac{C}{x + 1} + \frac{D}{x - 1}$
$1 = (A x + B) (x + 1) (x - 1) + C (x^{2} + 1) (x - 1) + D (x^{2} + 1) (x + 1)$
Put $x = 1, 1 = D (2) (2) \Rightarrow D = \frac{1}{4}$
Put $x = - 1, 1 = C (2) (- 2) \Rightarrow C = - \frac{1}{4}$
Put $x = 0, 1 = - B - C + D \Rightarrow B = - C + D - 1$
$\Rightarrow B = \frac{1}{4} + \frac{1}{4} - 1 =\Rightarrow B = - \frac{1}{2}$
Put $x = 2, 1 = (2 A + B) (3) (1) + C (5) (1) + D (5) (3)$
$\Rightarrow 1 = 6 a + 3 B + 5 C + 15 D$
$\Rightarrow 6 A = 1 + \frac{3}{2} + \frac{5}{4} - \frac{15}{4} = \frac{4 + 6 + 5 - 15}{4}$
$\Rightarrow 6 A = 0 \Rightarrow A = 0$
So

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