202203061946 Solution to 1967-CE-AMATH-I-XX

Find the indefinite integral $\int x\sqrt{x^2-2}\,\mathrm{d}x$ .

Solution.

$\begin{aligned} &\quad \int x\sqrt{x^2-2}\,\mathrm{d}x \\ &\enspace \cdots \enspace \cdots \enspace\cdots \\ \qquad & \textrm{let }y=x^2-2 \\ \qquad & \textrm{then }\mathrm{d}y=2x\,\mathrm{d}x \\ &\enspace \cdots \enspace\cdots \enspace\cdots \\ & = \int (x)(\sqrt{y})\bigg(\frac{\mathrm{d}y}{2x}\bigg) \\ & = \frac{1}{2}\int y^{\frac{1}{2}}\,\mathrm{d}y \\ & = \frac{y^{\frac{3}{2}}}{3} + C\textrm{ for some arbitrary constant }C\\ & = \frac{\sqrt{(x^2-2)^3}}{3} + C \\ \end{aligned}$

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