Problem 118
Hard Exponential and Logarithmic

Evaluate the integral: I = integral x/(x^4+3x²+2) dx

Evaluate the integral: $$ I = \int \frac{x}{x^4+3x^2+2}\,dx. $$

Explanation of Method

Factoring the quartic denominator as a product of two quadratics in $x^2$ allows partial fraction decomposition. Since only odd powers of $x$ appear in the numerator, each partial fraction term integrates directly via a $u=x^2$-style substitution to a logarithm.

  1. Factor the denominator.
  2. Decompose into partial fractions.
  3. Solve for the coefficients.
  4. Integrate each term and combine.

Step 1 · Factor the denominator

Treating $x^4+3x^2+2$ as a quadratic in $x^2$:

$$ x^4+3x^2+2 = (x^2+1)(x^2+2). $$

Thus, the integral becomes:

$$ I = \int \frac{x}{(x^2+1)(x^2+2)}\,dx. $$

Step 2 · Partial fraction decomposition

Assume:

$$ \frac{x}{(x^2+1)(x^2+2)} = \frac{Ax}{x^2+1} + \frac{Bx}{x^2+2}. $$

Multiplying both sides by $(x^2+1)(x^2+2)$ and dividing by $x$ (for $x\neq0$):

$$ 1 = A(x^2+2) + B(x^2+1). $$

Step 3 · Solve for the coefficients

Matching coefficients of $x^2$ and the constant term:

$$ A+B = 0, \qquad 2A+B = 1. $$

Subtracting the first from the second gives $A=1$, and so $B=-1$.

Step 4 · Integrate and combine

Substituting back:

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

Each term is a standard "numerator matches half the derivative of the denominator" pattern:

$$ I = \frac12\ln(x^2+1) - \frac12\ln(x^2+2) + C = \frac12\ln\left|\frac{x^2+1}{x^2+2}\right| + C. $$


Summary Table

Coefficient Value
$A$$1$
$B$$-1$
Antiderivative$\tfrac12\ln\left|\tfrac{x^2+1}{x^2+2}\right|$

Additional Notes

Result

The solution to the integral is:

$$ I = \frac12\ln\left|\frac{x^2+1}{x^2+2}\right| + C. $$