202203071350 Solution to 1968-CE-AMATH-I-XX

Find the indefinite integrals:

i. $\displaystyle{\int\frac{\mathrm{d}x}{x^4}}$

ii. $\displaystyle{\int\frac{\mathrm{d}x}{(2x-3)^4}}$

Solution.

i.

$\begin{aligned} &\quad \int\frac{\mathrm{d}x}{x^4} \\ & = \int x^{-4}\,\mathrm{d}x \\ & = \frac{x^{((-4)+1)}}{((-4)+1)}+C\textrm{ for some arbitrary constant }C \\ & = -\frac{1}{3x^3}+C \\ \end{aligned}$

ii.

$\begin{aligned} &\quad \int \frac{\mathrm{d}x}{(2x-3)^4} \\ & \cdots\enspace\cdots\enspace\cdots \\ & \textrm{let }y=2x-3\\ & \textrm{then }\frac{\mathrm{d}y}{\mathrm{d}x}=2 \\ & \textrm{and }\mathrm{d}x=\frac{\mathrm{d}y}{2} \\ & \cdots\enspace\cdots\enspace\cdots \\ & = \int\frac{\frac{\mathrm{d}y}{2}}{y^4} \\ & = \frac{1}{2}\int\frac{\mathrm{d}y}{y^4} \\ & \stackrel{\textrm{(i)}}{=} \frac{1}{2}\bigg( -\frac{1}{3x^3}+C\bigg) \\ & = -\frac{1}{6x^3}+C'\textrm{ for some arbitrary constant }C' \\ \end{aligned}$

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