Mathematics 9709/11 — October/November 2010

Walkthrough

This integral is not in a directly integrable form because of the squared bracket. We first expand:

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

The middle term is $2x \cdot \frac{1}{x} = 2$. Omitting this term is a common error.

Now integrate each term using the power rule $\int x^n\,dx = \frac{x^{n+1}}{n+1}$ for $n \neq -1$. For $x^{-2}$, $n = -2$, so:

$$\int x^{-2}\,dx = \frac{x^{-1}}{-1} = -\frac{1}{x}$$

Combining all terms and adding the constant $c$ gives the final answer. The mark scheme awards a mark for each correct term in the integrated expression, so even if one term is wrong, the other terms can still gain credit.

Key Takeaways

• Expand products before integrating when possible.
• Integrate power functions term by term.
• The reciprocal square $1/x^2$ integrates to $-1/x$, not to a logarithm.
• Always include the constant of integration for an indefinite integral.

Common Mistakes

• Forgetting the middle term $2$ when expanding the square.
• Treating $\int x^{-2}\,dx$ as $\ln x$; this is incorrect because $x^{-2}$ is a power function with exponent $-2$, not $x^{-1}$.
• Forgetting the constant of integration $c$.
• Writing $+ x^{-1}$ instead of $- x^{-1}$ after integrating.

Things to Be Careful About

• The original integrand is undefined at $x = 0$, so the result is valid on intervals not containing $0$; for an indefinite integral this is usually not penalised.
• Keep all three terms after integration: $x^3/3$, $2x$, and $-1/x$.
• The mark scheme states that omission of the middle term of the expansion can still earn 2 out of 3 marks, so show the expansion clearly.