AS 91579 · 2025 paper · substitution-type, constants, kinematics

Question 2(a) Achieved

A power-of-a-linear integral, a rate of change integrated back to its function, a constant found from a definite integral, a kinematics chain (acceleration → velocity → displacement), and an odd-power trig area.

Find int (10/(2x + 1)⁶) dx.

The 2025 paper, page 6 ↗ Marking schedule, page 3 ↗

Work it out

Try it on paper first. Stuck? Take one hint at a time.

1 Bring the bracket up from the bottom of the fraction.
10/(2x + 1)⁶ = 10(2x + 1)⁻⁶
2 Treat the bracket like x: add one to the power and divide by the new power. Then divide by the 2 inside the bracket.
int 10(2x + 1)⁻⁶ dx = (10(2x + 1)⁻⁵)/(−5 × 2) + c = −(2x + 1)⁻⁵ + c
3 Check it by differentiating your answer.
(d/dx)(−(2x + 1)⁻⁵) = 5(2x + 1)⁻⁶ × 2 = 10/(2x + 1)⁶

Answer −(2x + 1)⁻⁵ + c, which is −1/(2x + 1)⁵ + c.

What the marker wanted

Achieved The correct answer, in either form.

In our words, from NZQA’s 2025 marking schedule.

Common mistake

Writing a logarithm because there is a bracket on the bottom. NZQA’s 2025 report lists turning any fraction with f(x) underneath into ln f(x) among the Not Achieved answers. A log comes only from a power of −1; this bracket is to the power −6.

How did it go?

Got it moves it further out. Review brings it back tomorrow.