Integration Techniques: \(e^{|x-1|}\); trig identity; \(x=u^6\) — EJC 2025 H2 Math Prelim Paper 1

Split at \(x=1\) (where \(|x-1|\) changes): \[\begin{aligned} \int_0^4 \mathrm{e}^{|x-1|}\,\mathrm{d}x &= \int_0^1 \mathrm{e}^{1-x}\,\mathrm{d}x + \int_1^4 \mathrm{e}^{x-1}\,\mathrm{d}x\\[4pt] &= \Bigl[-\mathrm{e}^{1-x}\Bigr]_0^1 + \Bigl[\mathrm{e}^{x-1}\Bigr]_1^4\\[4pt] &= \bigl(-1 + \mathrm{e}\bigr) + \bigl(\mathrm{e}^3 - 1\bigr)\\[4pt] &= \mathrm{e}^3 + \mathrm{e} - 2 \end{aligned}\]

\(\mathrm{e}^3 + \mathrm{e} - 2\) \[\begin{aligned} \int \sin^2 3x + \tan^2 3x \, \mathrm{d}x &= \int \frac{1-\cos 6x}{2} + \sec^2 3x - 1 \, \mathrm{d}x \\ &= \int -\frac{1}{2}\cos 6x + \sec^2 3x - \frac{1}{2} \, \mathrm{d}x \\ &= -\frac{1}{12}\sin 6x + \frac{1}{3}\tan 3x - \frac{1}{2}x + C \end{aligned}\] \(-\dfrac{1}{12}\sin 6x + \dfrac{1}{3}\tan 3x - \dfrac{1}{2}x + C\)

Since \(\dfrac{\mathrm{d}x}{\mathrm{d}u} = 6u^5\),

\[\begin{aligned} \int \frac{1}{\sqrt{x}+\sqrt[3]{x}}\,\mathrm{d}x &= \int \frac{1}{u^3+u^2} \cdot 6u^5\,\mathrm{d}u \\ &= 6\int \frac{u^3}{u+1}\,\mathrm{d}u \\ &= 6\int u^2 - u + 1 - \frac{1}{u+1}\,\mathrm{d}u \\ &= 6\left[\frac{1}{3}u^3 - \frac{1}{2}u^2 + u - \ln(u+1)\right] + C \\ &= 2\sqrt{x} - 3\sqrt[3]{x} + 6\sqrt[6]{x} - 6\ln\!\left(\sqrt[6]{x}+1\right) + C \end{aligned}\]