(i)[4]
With the substitution $u = \sqrt{6 - x}$, demonstrate that $I = \int_{1}^{2} \frac{10u}{(3 - u)(2 + u)} \, du$.
(ii)[6]
Hence show that $I = 2\ln\left(\frac{9}{2}\right)$.
Mathematics 9709 · AS & A Level · Integration
Integration — practice question
Take $I = \int_{2}^{5} \frac{5}{x + \sqrt{6 - x}} \, dx$.