Specialist2024 NHT Exam 2 Section A · Q8CAS active1 mark

Substitution in a definite integral

Using the substitution $u^2 = x + 1$ , the definite integral $\displaystyle\int_3^8 \left(\frac{1}{x\sqrt{x+1}}\right) dx$ can be expressed as

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\displaystyle\frac{1}{2}\int_2^3 \left(\frac{1}{u-1} - \frac{1}{u+1}\right) du

\displaystyle\int_2^3 \left(\frac{1}{u-1} - \frac{1}{u+1}\right) du

\displaystyle\int_2^3 \left(\frac{1}{u+1} - \frac{1}{u-1}\right) du

\displaystyle\int_3^8 \left(\frac{1}{u+1} - \frac{1}{u-1}\right) du

\displaystyle\int_2^3 \left(\frac{1}{u-1} + \frac{1}{u+1} - \frac{2}{u}\right) du

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