Q.Evaluate: $\int \dfrac{x^2+2}{x+1}\,dx$

The key idea is Partial Fraction Decomposition — but here the rational function is improper (numerator degree ≥ denominator degree), so we must first divide.

Step 1: Perform polynomial division.

Divide $x^2+2$ by $x+1$:

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

Step 2: Rewrite the integrand.

Step 3: Integrate term by term.

When you see a rational function where the numerator’s degree is greater than or equal to the denominator’s degree, your first instinct should be: divide first. The fraction $\frac{x^2+2}{x+1}$ has a quadratic numerator and a linear denominator — the degree of the numerator (2) is higher than that of the denominator (1). That means the fraction is “improper” in the rational-function sense. You can’t directly apply partial fractions until you’ve extracted the polynomial part.

Why does this work? Think of it like converting an improper fraction in arithmetic, say $\frac{7}{3}$: you write it as $2 + \frac{1}{3}$ because $7 \div 3$ gives quotient 2 and remainder 1. Here, we do the same with polynomials: divide $x^2+2$ by $x+1$ to get a quotient (a polynomial) and a remainder (a proper fraction). The remainder will have a lower degree than the denominator, and that part can be integrated using a simple log.

Let’s do it step by step.

1. Perform polynomial long division

Divide $x^2 + 0x + 2$ by $x+1$:

• $x^2 \div x = x$. Multiply: $x(x+1) = x^2 + x$. Subtract from $x^2 + 2$: $(x^2 + 2) - (x^2 + x) = -x + 2$.
• Now divide $-x \div x = -1$. Multiply: $-1(x+1) = -x - 1$. Subtract: $(-x + 2) - (-x - 1) = 3$.

So the quotient is $x - 1$ and the remainder is $3$. Therefore:

2. Rewrite the integral

Now the integral becomes:

This splits into three simple integrals:

3. Integrate each term

• $\int x\,dx = \frac{x^2}{2}$
• $\int 1\,dx = x$
• $\int \frac{1}{x+1}\,dx = \log|x+1|$ (don’t forget the absolute value — the denominator could be negative for $x < -1$)

So:

4. Check by differentiating (optional but good practice)

Differentiate $\frac{x^2}{2} - x + 3\log|x+1| + C$:

• Derivative of $\frac{x^2}{2}$ is $x$
• Derivative of $-x$ is $-1$
• Derivative of $3\log|x+1|$ is $\frac{3}{x+1}$

Sum: $x - 1 + \frac{3}{x+1}$, which is exactly the original integrand. So it’s correct.

Method: Integrating an improper rational function by long division

Use this whenever $\int\dfrac{P(x)}{Q(x)}\,dx$ has $\deg P \ge \deg Q$. You must reduce it to a polynomial plus a proper fraction before integrating.

Steps

Step 1: Check the degrees.

If the numerator's degree is greater than or equal to the denominator's, the fraction is improper and cannot be integrated term-by-term as it stands.

Step 2: Do polynomial long division.

Divide to write

e.g. $\dfrac{x^2+2}{x+1}=(x-1)+\dfrac{3}{x+1}$, where the remainder's degree is now less than $Q$'s.

Step 3: Integrate each piece with standard rules.

The polynomial part uses the power rule; the proper fraction $\dfrac{c}{x+a}$ integrates to $c\log|x+a|$.

Step 4: Combine and add $C$.

Assemble the results, e.g. $\dfrac{x^2}{2}-x+3\log|x+1|+C$. The key recognition: a top-heavy fraction is a signal to divide first, never to jump straight to $\log$ or partial fractions.

Common Mistakes

Mistake 1: Integrating the improper fraction directly.

Why it's wrong: $\dfrac{x^2+2}{x+1}$ is not $\log$-integrable as written because the top degree exceeds the bottom's. Correct approach: perform long division first to split off a polynomial part.

Mistake 2: Getting the remainder wrong.

Why it's wrong: $x^2+2=(x+1)(x-1)+3$, so the remainder is $3$; an incorrect remainder changes the $\log$ coefficient. Correct approach: verify by expanding $(x+1)(x-1)+3=x^2+2$.

Mistake 3: Forgetting the constant coefficient on the log.

Why it's wrong: $\int\tfrac{3}{x+1}\,dx=3\log|x+1|$, not $\log|x+1|$ — the $3$ from the remainder must be carried. Correct approach: integrate $\tfrac{\text{remainder}}{x+1}$ keeping its numerical coefficient.