Specialist2026 NHT Exam 2 Section A · Q9CAS active1 mark

Substitution in a definite integral

Using the substitution $x^3 - 1 = u^2$ , $u \geq 0$ , the definite integral $\displaystyle\int_1^2 \frac{\sqrt{x^3 - 1}}{x}\,dx$ can be expressed as

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\displaystyle\int_0^{\sqrt{7}} \frac{2}{3}\left(1 - \frac{1}{1+u^2}\right)du

\displaystyle\int_0^{\sqrt{7}} \frac{1}{3}\left(\frac{u}{1+u^2}\right)du

\displaystyle\int_1^{2} \frac{3}{2}\left(1 + \frac{1}{1-u^2}\right)du

\displaystyle\int_1^{2} \frac{2}{3}\left(1 + \frac{1}{1+u^2}\right)du

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