Question

express the integrand as a sum of partial fractions and then evaluate the integral.
int\frac{10x^{2}}{x^{4}-1}dx
express the integrand as a sum of partial fractions.
\frac{10x^{2}}{x^{4}-1}=square
(simplify your answer. use integers or fractions for any numbers in the expression.)

Explanation:

Step1: Factor the denominator

$x^{4}-1=(x^{2} + 1)(x^{2}-1)=(x^{2}+1)(x + 1)(x - 1)$

Step2: Set up partial - fraction decomposition

$\frac{10x^{2}}{x^{4}-1}=\frac{Ax + B}{x^{2}+1}+\frac{C}{x + 1}+\frac{D}{x - 1}$
$10x^{2}=(Ax + B)(x + 1)(x - 1)+C(x^{2}+1)(x - 1)+D(x^{2}+1)(x + 1)$

Step3: Find the values of A, B, C, and D

Let $x = 1$:
$10\times1^{2}=D(1^{2}+1)(1 + 1)$
$10 = 4D$, so $D=\frac{5}{2}$
Let $x=-1$:
$10\times(-1)^{2}=C((-1)^{2}+1)(-1 - 1)$
$10=-4C$, so $C =-\frac{5}{2}$
Expand the right - hand side:
$(Ax + B)(x^{2}-1)+C(x^{3}-x^{2}+x - 1)+D(x^{3}+x^{2}+x + 1)$
$=(Ax + B)(x^{2}-1)+(C + D)x^{3}+(D - C)x^{2}+(C + D)x+(D - C)$
Since the coefficient of $x^{3}$ is 0, $A + C+D = 0$. Substituting $C =-\frac{5}{2}$ and $D=\frac{5}{2}$, we get $A = 0$.
Since the coefficient of $x$ is 0, $C + D=0$ (which is already satisfied as $C=-\frac{5}{2}$ and $D = \frac{5}{2}$), and substituting $A = 0$ into $10x^{2}=(Ax + B)(x + 1)(x - 1)+C(x^{2}+1)(x - 1)+D(x^{2}+1)(x + 1)$ and comparing the constant terms and $x^{2}$ terms, we find $B = 5$.
So $\frac{10x^{2}}{x^{4}-1}=\frac{5}{x^{2}+1}-\frac{5/2}{x + 1}+\frac{5/2}{x - 1}$

Answer:

$\frac{5}{x^{2}+1}-\frac{5}{2(x + 1)}+\frac{5}{2(x - 1)}$