Which expression is equal to $$\int \frac{1}{x^2 + 4x + 10} \, dx ?$$ A - HSC - SSCE Mathematics Extension 2 - Question 9 - 2020 - Paper 1

The expression in the denominator can be rewritten as follows: $x^2 + 4x + 10 = (x^2 + 4x + 4) + 6 = (x + 2)^2 + 6.$ Thus, the integral simplifies to: $\int \frac{1}{(x + 2)^2 + 6} \, dx.$

This expression resembles the standard form for the integral of the arctangent function. We can use the substitution: $a = \sqrt{6}, \quad u = x + 2.$ The integral then becomes: $\int \frac{1}{u^2 + a^2} \, du = \frac{1}{a} \tan^{-1}\left( \frac{u}{a} \right) + c.$ Substituting back our expressions gives: $\frac{1}{\sqrt{6}} \tan^{-1}\left( \frac{x + 2}{\sqrt{6}} \right) + c.$

Finally, we compare our result with the given options. Our derived expression matches with: A. ( \frac{1}{\sqrt{6}} \tan^{-1}\left( \frac{x + 2}{\sqrt{6}} \right) + c ) Thus, the correct answer is A.