Question

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

• A. log $x(x^2-1)+c$, where c is a constant of integration.
• B. log $(\frac{x^2-1}{x})+c$, where c is a constant of integration.
• C. log$(x^2-1)+c$, where c is a constant of integration.
• D. log$(\frac{x^2+1}{x})+c$, where c is a constant of integration.

Answer 1

Answer: (B)

log $(\frac{x^2-1}{x})+c$, where c is a constant of integration.

Answer 2

The integral we need to solve is $\int \frac{x^2+1}{x(x^2-1)} dx$. The denominator can be factored as $x(x-1)(x+1)$. We can use partial fraction decomposition for the integrand:

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

Multiplying both sides by $x(x-1)(x+1)$:

$x^2+1 = A(x-1)(x+1) + Bx(x+1) + Cx(x-1)$

Substituting the roots of the denominator:

$x=0 \implies 1 = A(-1)(1) \implies A = -1$

$x=1 \implies 2 = B(1)(2) \implies B = 1$

$x=-1 \implies 2 = C(-1)(-2) \implies C = 1$

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

Integrating each term:

$\int \frac{x^2+1}{x(x^2-1)} dx = \int \left(\frac{-1}{x} + \frac{1}{x-1} + \frac{1}{x+1}\right) dx = -\log|x| + \log|x-1| + \log|x+1| + C$

Using logarithm properties:

$-\log|x| + \log|x-1| + \log|x+1| = \log\left|\frac{x^2-1}{x}\right| + C$