We need to evaluate the definite integral $J = \int \frac{x^2+x-2}{x^4+5x^2+4} dx$.

1. Problem Description

We need to evaluate the definite integral

$J = \int \frac{x^2+x-2}{x^4+5x^2+4} dx$.

2. Solution Steps

First, we factor the denominator:

$x^4 + 5x^2 + 4 = (x^2+1)(x^2+4)$.

Then, we factor the numerator:

$x^2+x-2 = (x+2)(x-1)$.

So, we have

$J = \int \frac{(x+2)(x-1)}{(x^2+1)(x^2+4)} dx = \int \frac{x^2+x-2}{(x^2+1)(x^2+4)} dx$.

Next, we perform partial fraction decomposition:

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

Multiplying both sides by $(x^2+1)(x^2+4)$, we get:

$x^2+x-2 = (Ax+B)(x^2+4) + (Cx+D)(x^2+1)$.

$x^2+x-2 = Ax^3 + 4Ax + Bx^2 + 4B + Cx^3 + Cx + Dx^2 + D$.

$x^2+x-2 = (A+C)x^3 + (B+D)x^2 + (4A+C)x + (4B+D)$.

Equating coefficients, we have the following system of equations:

$A+C = 0$

$B+D = 1$

$4A+C = 1$

$4B+D = -2$

From the first equation, $C = -A$.

Substituting into the third equation, $4A - A = 1$, so $3A = 1$, and $A = \frac{1}{3}$.

Thus, $C = -\frac{1}{3}$.

From the second equation, $D = 1-B$.

Substituting into the fourth equation, $4B + 1 - B = -2$, so $3B = -3$, and $B = -1$.

Thus, $D = 1 - (-1) = 2$.

So, we have $A = \frac{1}{3}, B = -1, C = -\frac{1}{3}, D = 2$.

Thus,

$\frac{x^2+x-2}{(x^2+1)(x^2+4)} = \frac{\frac{1}{3}x-1}{x^2+1} + \frac{-\frac{1}{3}x+2}{x^2+4}$.

$J = \int \left( \frac{\frac{1}{3}x}{x^2+1} - \frac{1}{x^2+1} - \frac{\frac{1}{3}x}{x^2+4} + \frac{2}{x^2+4} \right) dx$

$J = \frac{1}{3} \int \frac{x}{x^2+1} dx - \int \frac{1}{x^2+1} dx - \frac{1}{3} \int \frac{x}{x^2+4} dx + 2 \int \frac{1}{x^2+4} dx$

$J = \frac{1}{6} \ln(x^2+1) - \arctan(x) - \frac{1}{6} \ln(x^2+4) + \arctan(\frac{x}{2}) + C$.

3. Final Answer

$J = \frac{1}{6} \ln(x^2+1) - \arctan(x) - \frac{1}{6} \ln(x^2+4) + \arctan(\frac{x}{2}) + C$

$J = \frac{1}{6}\ln\left(\frac{x^2+1}{x^2+4}\right) - \arctan(x) + \arctan(\frac{x}{2}) + C$