Joshua Foley

Answered

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.

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.

Answer & Explanation

Zackery Harvey

Expert

2022-07-02Added 21 answers

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 ${I}_{f}(x)=\int {f}^{-1}(x)dx.$.

We set $x=f(u)$ to get ${I}_{f}(x)=\int u{f}^{\prime}(u)du.$.

Step 2

Integration by parts gives ${I}_{f}(x)=uf(u)-\int f(u)du$ which is ${I}_{f}(x)=x{f}^{-1}(x)-F\circ {f}^{-1}(x)+C.$.

Choosing ${f}^{-1}(x)=\mathrm{ln}x,$, ${I}_{\mathrm{ln}}(x)=x\mathrm{ln}x-x+C.$.

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 ${I}_{f}(x)=\int {f}^{-1}(x)dx.$.

We set $x=f(u)$ to get ${I}_{f}(x)=\int u{f}^{\prime}(u)du.$.

Step 2

Integration by parts gives ${I}_{f}(x)=uf(u)-\int f(u)du$ which is ${I}_{f}(x)=x{f}^{-1}(x)-F\circ {f}^{-1}(x)+C.$.

Choosing ${f}^{-1}(x)=\mathrm{ln}x,$, ${I}_{\mathrm{ln}}(x)=x\mathrm{ln}x-x+C.$.

Janet Forbes

Expert

2022-07-03Added 4 answers

Explanation:

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

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

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