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I have a problem with this integral. $$\int\limits_{0}^{(e-1)^2}\!\! \left({\ln(\sqrt{x}+1)} \right)\,\mathrm dx $$ I applied the substitution method $t = \sqrt{x}+1$, $2t = dx$

I changed integration interval from $0 \to (e-1)^2$ to $1 \to e$

$$\int\limits_{1}^{e}\!\! \left(2t\,{\ln(t}) \right)\,\mathrm dt $$ Then I worked for integration by parts $$={t^2 \ln t}-\int\limits \left (t \right)\,\mathrm dt$$ $$={t^2 \ln t-\frac{t^2}{2}+C} $$

completing the exercise I understand the following result $\frac{e^2+1}{2}$ , which it is obviously wrong.

Can you correct my mistake?

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  • $\begingroup$ You made a mistake differentiating your substitution $\endgroup$ – David Quinn Dec 30 '15 at 16:01
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    $\begingroup$ I think you should have $dx=2(t-1)dt$. $\endgroup$ – kccu Dec 30 '15 at 16:04
  • $\begingroup$ @zz20s. $\log$ is most often used for base 10 logarithm, while $\ln$ is used for the base e logarithm. $\endgroup$ – wythagoras Dec 30 '15 at 16:26
  • $\begingroup$ @zz20s While I much prefer “log” to “ln”, the latter is unfortunately common, so it's better to leave the notation used by the OP. I heartily disagree with those who think that “log” means base 10; no, in mathematics it means the only useful logarithm, that is, base $e$. $\endgroup$ – egreg Dec 30 '15 at 16:44
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There was an error in the substitution: $$t=\sqrt{x}+1 \implies dt =\frac{1}{2\sqrt{x}}dx=\frac{1}{2(\sqrt{x}+1)-2}dx=\frac{1}{2t-2}dx \implies dx=2(t-1)dt$$ So $$\int_{1}^e2(t-1)\ln(t)\,dt=\int_{1}^e\left((t-1)^2\right)'\ln(t)\,dt=[(t-1)^2\ln(t)]_{1}^{e}-\int_{1}^e\frac{(t-1)^2}{t}\,dt$$ I assume you can take it from here.

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  • $\begingroup$ I get it. Thanks :) $\endgroup$ – user12 Dec 30 '15 at 16:12
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First use the substitution $t=\sqrt x$. This gives $dt=\frac{1}{2}x^{-\frac{1}{2}}dx$, or $dx=2\sqrt x\ dt=2t\ dt$.

Re-evaluate the limits next. When $x=0$, $t=0$, and when $x=(e-1)^2$, $t=e-1$.

The new integral is

$$\int_0^{e-1} 2t \ln(t+1)dt$$

Finally we integrate by parts: $u=\ln(t+1)$ and $dv=2tdt$. Then $du=\frac{1}{t+1}dt$ and $v=t^2$.

Now we have

$$t^2\ln(t+1)+\int\frac{t^2}{t+1}dt$$

You can integrate by parts again, etc.

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In order to solve $$\int_0^{(e-1)^2}{(\ln(\sqrt{x}+1))\mathrm{d}x},$$ We will now use the substitution $\mathrm{u}=\sqrt{x}+1$, or $x=(\mathrm{u}-1)^2$, which provides $\mathrm{d}x = 2(\mathrm{u}-1)\mathrm{du}$. From this, we obtain $$\int_0^{(e-1)^2}{(2(\mathrm{u}-1)\ln(\mathrm{u}))\mathrm{du}}.$$ Using integration by parts with $g'(\mathrm{u})=\ln(\mathrm{u})$ and $h(\mathrm{u})=2(\mathrm{u}-1)$, we can calculate $g(\mathrm{u})=\mathrm{u}\ln(\mathrm{u})-\mathrm{u}$ and $h'(\mathrm{u})=2$. Now, we have $$\int_0^{(e-1)^2}{(2(\mathrm{u}-1)\ln(\mathrm{u}))\mathrm{du}} = \left. 2(\mathrm{u}-1)(\mathrm{u}\ln(\mathrm{u})-\mathrm{u})\right\vert_{0}^{(e-1)^2}-\int_0^{(e-1)^2}{(2(\mathrm{u}\ln(\mathrm{u})-\mathrm{u}))\mathrm{du}}.$$ Wait, something seems familiar. See it? The second term in the right hand side is the same as the integral we're trying to calculate. We'll use this to our advantage by substituting $y$ is equal to that big nasty integral. Our equation is now $$y=\left. 2(\mathrm{u}-1)(\mathrm{u}\ln(\mathrm{u})-\mathrm{u}) \right\vert_{0}^{(e-1)^2} -y.$$ Adding $y$ to both sides, our equation is now $$2y=\left. 2(\mathrm{u}-1)(\mathrm{u}\ln(\mathrm{u})-\mathrm{u}) \right\vert_{0}^{(e-1)^2} ,$$ which is the same as $$y=\left. (\mathrm{u}-1)(\mathrm{u}\ln(\mathrm{u})-\mathrm{u}) \right\vert_{0}^{(e-1)^2}.$$ I think you've got this from here. Just substitute and simplify! ;-)

If I've made some errors, somebody please point them out.

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