Q.Find: $\displaystyle\int \dfrac{x^2 + 1}{(x^2 + 2)(x^2 + 4)}\, dx$

Concept: Integration by partial fractions after a substitution to simplify the denominator.

We rewrite the integral by splitting the numerator using the identity $x^2 + 1 = A(x^2 + 2) + B(x^2 + 4)$. Solving for $A$ and $B$:

Comparing coefficients: $A + B = 1$ and $2A + 4B = 1$.

Subtracting twice the first equation from the second gives $2B = -1$, so $B = -\frac12$, and then $A = \frac32$.

Thus the integral becomes:

$\int \frac{3/2}{x^2 + 2} \, dx - \int \frac{1/2}{x^2 + 4} \, dx$ …

The integrand is a rational function where both numerator and denominator are polynomials in $x^2$. The denominator factors as $(x^2+2)(x^2+4)$, so the natural approach is partial fractions — but with a twist. Since every term is even in $x$, we can treat $t = x^2$ as a variable, decompose in $t$, then integrate each resulting term using standard inverse tangent forms.

The key insight: when the integrand is of the form $\frac{P(x^2)}{Q(x^2)}$ with denominator factored into distinct quadratic factors, the partial fraction decomposition in terms of $x^2$ works cleanly. Each term will be of the form $\frac{A}{x^2 + a^2}$, whose integral is $\frac{1}{a}\tan^{-1}\frac{x}{a} + C$.

Let's work through it.

1. Set up the substitution. Let $t = x^2$. Then the integrand becomes $\frac{t+1}{(t+2)(t+4)}$. We want constants $A$ and $B$ such that:

2. Solve for $A$ and $B$. Multiply both sides by $(t+2)(t+4)$:

Expand: $t+1 = (A+B)t + (4A + 2B)$.

Comparing coefficients:

From the first, $B = 1 - A$. Substitute into the second:

Then $B = 1 - (-\frac12) = \frac32$.

So:

3. Rewrite in terms of $x$. Substituting back $t = x^2$:

4. Integrate term by term. Recall $\int \frac{dx}{x^2 + a^2} = \frac{1}{a}\tan^{-1}\frac{x}{a} + C$.

For the first integral, $a = \sqrt{2}$; for the second, $a = 2$.