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

Concept: Integration by partial fractions — the denominator is already factored, so we decompose the rational function into simpler fractions.

Step 1 – Set up the partial fraction decomposition.

Since the denominator has a linear factor $(x+2)$ and an irreducible quadratic $(x^2+1)$, write:

Step 2 – Solve for $A, B, C$.

Multiply through by $(x+2)(x^2+1)$:

Expand and collect:

Comparing coefficients:

• $x^2$: $A + B = 1$
• $x$: $2B + C = 1$
• constant: $A + 2C = 1$ …

The key here is that the integrand is a rational function — a polynomial divided by a polynomial. When the denominator factors into linear and irreducible quadratic factors, partial fractions are the standard tool. The numerator is of degree 2, the denominator of degree 3, so the fraction is proper (no polynomial division needed). We break it into simpler pieces that integrate directly to logs and arctangents.

Let’s work through it.

1. Set up the partial fraction decomposition. The denominator is $(x+2)(x^2+1)$. The factor $x+2$ is linear, and $x^2+1$ is an irreducible quadratic (no real roots). So we write:

where $A$, $B$, $C$ are constants to find.

2. Clear denominators. Multiply both sides by $(x+2)(x^2+1)$:

3. Expand and collect like terms.

Adding them:

4. Equate coefficients. Comparing coefficients of $x^2$, $x$, and the constant term gives the system:

5. Solve the system.

   From the first equation, $B = 1 - A$. Substitute into the second: $2(1 - A) + C = 1 \implies 2 - 2A + C = 1 \implies C = 2A - 1$.

   Substitute into the third: $A + 2(2A - 1) = 1 \implies A + 4A - 2 = 1 \implies 5A = 3 \implies A = \frac{3}{5}$.

   Then $B = 1 - \frac{3}{5} = \frac{2}{5}$, and $C = 2\cdot\frac{3}{5} - 1 = \frac{6}{5} - 1 = \frac{1}{5}$.

   So:

6. Integrate term by term.

For the second term, split it:

Now integrate each: …