Question

Evaluate: $\int \frac{x^{2}}{x^{4}+x^{2}-2} ,dx$

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Answer


To evaluate the integral $$ \int \frac{x^{2}}{x^{4}+x^{2}-2} , dx $$ we start by simplifying the integrand.

Simplifying the integrand:

We begin by factoring the expression $x^4 + x^2 - 2$ in the denominator: $$ x^4 + x^2 - 2 = (x^2 - 1)(x^2 + 2). $$ Thus, the integral becomes: $$ \int \frac{x^{2}}{(x^2 - 1)(x^2 + 2)} , dx. $$

Next, we factor $x^2 - 1$ further: $$ (x^2 - 1) = (x - 1)(x + 1). $$ So the integrand can be written as: $$ \int \frac{x^{2}}{(x - 1)(x + 1)(x^2 + 2)} , dx. $$

Using partial fractions:

To separate the integrand into simpler pieces, we use partial fraction decomposition: $$ \frac{x^{2}}{(x - 1)(x + 1)(x^2 + 2)} = \frac{A}{x - 1} + \frac{B}{x + 1} + \frac{C x + D}{x^2 + 2}. $$ Clearing the denominators, we get: $$ x^{2} = A(x + 1)(x^2 + 2) + B(x^2 + 2)(x - 1) + (C x + D)(x - 1)(x + 1). $$ By expanding and equating the coefficients of corresponding powers of $x$, we get a system of equations: $$ \begin{aligned} A + B + C &= 0, \ A - B + D &= 1, \ 2A + 2B - C &= 0, \ 2A - 2B - D &= 0. \end{aligned} $$ Solving this system, we find: $$ A = \frac{1}{6}, \quad B = -\frac{1}{6}, \quad C = 0, \quad D = \frac{2}{3}. $$

Rewriting the integral:

Substituting the values of $A$, $B$, $C$, and $D$ back into the partial fractions, we get: $$ \int \frac{x^{2}}{(x-1)(x+1)(x^2 + 2)} , dx = \int \left( \frac{1}{6(x-1)} - \frac{1}{6(x+1)} + \frac{2}{3(x^2 + 2)} \right) , dx. $$

Evaluating the integral:

We can now integrate each term separately: $$ \int \frac{1}{6(x-1)} , dx - \int \frac{1}{6(x+1)} , dx + \int \frac{2}{3(x^2 + 2)} , dx. $$ These integrals are: $$ \frac{1}{6} \ln|x - 1| - \frac{1}{6} \ln|x + 1| + \frac{2}{3} \left( \frac{1}{\sqrt{2}} \tan^{-1} \left( \frac{x}{\sqrt{2}} \right) \right) + C. $$

Simplifying, we have: $$ \frac{1}{6} \left( \ln|x - 1| - \ln|x + 1| + 2\sqrt{2} \tan^{-1} \left( \frac{x}{\sqrt{2}} \right) \right) + C. $$

Thus, the evaluated integral is: $$ \int \frac{x^{2}}{x^{4}+x^{2}-2} , dx = \frac{1}{6} \left( \ln|x - 1| - \ln|x + 1| + 2\sqrt{2} \tan^{-1} \left( \frac{x}{\sqrt{2}} \right) \right) + C. $$


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