Joshua Foley

2022-07-01

Antiderivative of log(x) without Parts
I understand how the antiderivative of log(x) can be obtained by Integration by Parts (i.e. product rule), but I was wondering how-if at all- it could be obtained only using sum/difference rule and substitution/chain rule.

Zackery Harvey

Expert

Step 1
This may not necessarily be what you're looking for, but I got a kick out of it so I figured I'd share. We may evaluate the antiderivative by using integration by parts indirectly.
We seek the integral ${I}_{f}\left(x\right)=\int {f}^{-1}\left(x\right)dx.$.
We set $x=f\left(u\right)$ to get ${I}_{f}\left(x\right)=\int u{f}^{\prime }\left(u\right)du.$.
Step 2
Integration by parts gives ${I}_{f}\left(x\right)=uf\left(u\right)-\int f\left(u\right)du$ which is ${I}_{f}\left(x\right)=x{f}^{-1}\left(x\right)-F\circ {f}^{-1}\left(x\right)+C.$.
Choosing ${f}^{-1}\left(x\right)=\mathrm{ln}x,$, ${I}_{\mathrm{ln}}\left(x\right)=x\mathrm{ln}x-x+C.$.

Janet Forbes

Expert

Explanation:
${\int }_{1}^{t}\mathrm{ln}\left(x\right)\phantom{\rule{thinmathspace}{0ex}}dx={\int }_{1}^{t}{\int }_{1}^{x}\frac{1}{u}\phantom{\rule{thinmathspace}{0ex}}du\phantom{\rule{thinmathspace}{0ex}}dx={\int }_{1}^{t}{\int }_{u}^{t}\frac{1}{u}\phantom{\rule{thinmathspace}{0ex}}dx\phantom{\rule{thinmathspace}{0ex}}du={\int }_{1}^{t}\frac{t-u}{u}\phantom{\rule{thinmathspace}{0ex}}du=\left[t\mathrm{ln}\left(u\right)-u{\right]}_{1}^{t}$

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