Answer: ∫(x²+1)/(x⁴-1)dx is equal to
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note that, by difference of squares, x⁴-1 = (x²+1)(x²-1), so
(x^2-1)} dx)
(x²+1) cancel so we are left with
(x-1)} dx)
we can break down
(x-1)})
into partial fractionssince there are two linear factors in the denominator, x + 1 and x - 1, we have the following partial fraction decomposition:
(x-1)} = \frac{a}{x+1} + \frac{b}{x-1})
we are trying to make an identity because we want to change the form of
into a decomposed partial fraction form. therefore, this equation is true for all values of x.to find the values of a and b, we can start by multiplying both sides of the equation by the lowest common denominator, (x+1)(x-1).
(x-1) \cdot \frac{1}{(x+1)(x-1)} & = \left(\frac{a}{x+1} + \frac{b}{x-1}\right) \cdot (x+1)(x-1) \\ 1 & = a(x-1) + b(x+1) \end{aligned})
we can use a shortcut for linear factors. since this equation is an identity and true for all values of x, we can use any value of x to us try to find the values of a and b that will establish this identity.if we choose x = 1, we can eliminate a and solve for b:
 + b(1+1) \\ 1 & = 0 + 2b \\ b & = 1/2 \end{aligned})
if we choose x = -1, we can eliminate b and solve for a
 + b(-1+1) \\ 1 & = -2a + 0 \\ a & = -1/2 \end{aligned})
therefore:
(x-1)} & = \frac{-\frac{1}{2}}{x+1} + \frac{\frac{1}{2}}{x-1}\end{aligned})
and we have
(x-1)} dx \\ \\ & = \int \left(\frac{-\frac{1}{2}}{x+1} + \frac{\frac{1}{2}}{x-1} \right) dx \\ \\ & = \int \frac{-\frac{1}{2}}{x+1} dx + \int \frac{\frac{1}{2}}{x-1}dx & & \text{\footnotesize (integral sum rule)} \\ \\ & = -\frac{1}{2}\int \frac{1}{x+1}dx + \frac{1}{2}\int \frac{1}{x-1}dx& & \text{\footnotesize (integral constant rule)} \\ \\ & = -\frac{1}{2} \ln|x+1| + \frac{1}{2}\ln|x-1| + c & & \text{\footnotesize ($u$-substitution)} \end{aligned})
