How do you use partial fractions to find the integral $\int \frac{1}{x^{2} - 1} d x$ $int 1/(x^2-1)dx$ ?

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How do you use partial fractions to find the integral $\int \frac{1}{x^{2} - 1} d x$ $int 1/(x^2-1)dx$ ?

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Answer 1 · 2016-12-09T06:28:11

The answer is $= = - \frac{1}{2} ln (∣ x + 1 ∣) + \frac{1}{2} ln (∣ x - 1 ∣) + C$ $==-1/2ln(∣x+1∣)+1/2ln(∣x-1∣)+C$

Explanation:

We use $a^{2} - b^{2} = (a + b) (a - b)$ $a^2-b^2=(a+b)(a-b)$

The denominator is

$(x^{2} - 1) = (x + 1) (x - 1)$ $(x^2-1)=(x+1)(x-1)$

So, the decompostion into partial fractions is

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

$= \frac{A (x - 1) + B (x + 1)}{(x + 1) (x - 1)}$ $=(A(x-1)+B(x+1))/((x+1)(x-1))$

Therefore,

$1 = A (x - 1) + B (x + 1)$ $1=A(x-1)+B(x+1)$

Let $x = 1$ $x=1$ , $\Rightarrow$ $=>$ , $1 = 2 B$ $1=2B$

Let $x = - 1$ $x=-1$ , $\Rightarrow$ $=>$ , $1 = - 2 A$ $1=-2A$

so,

$\frac{1}{x^{2} - 1} = - \frac{1}{2 (x + 1)} + \frac{1}{2 (x - 1)}$ $1/(x^2-1)=-1/(2(x+1))+1/(2(x-1))$

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

$= - \frac{1}{2} ln (∣ x + 1 ∣) + \frac{1}{2} ln (∣ x - 1 ∣) + C$ $=-1/2ln(∣x+1∣)+1/2ln(∣x-1∣)+C$

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How do you use partial fractions to find the integral $\int \frac{1}{x^{2} - 1} d x$ $int 1/(x^2-1)dx$ ?

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Explanation:

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