piarepm

2022-01-03

Finding integral:

${\int}_{0}^{1}\frac{\mathrm{ln}(1+x)}{x}dx$

Paul Mitchell

Beginner2022-01-04Added 40 answers

Here is an elementary integration of $I={\int}_{0}^{1}\frac{\mathrm{ln}(1+x)}{x}dx$

$I=-\frac{1}{2}{\int}_{0}^{1}\frac{\mathrm{ln}(1+x)}{x}dx+\frac{3}{2}{\int}_{0}^{1}\frac{\mathrm{ln}(1+x)}{x}dx=-\frac{3}{2}{\int}_{0}^{1}\frac{\mathrm{ln}(1-x+{x}^{2})}{x}dx$

Let$J\left(a\right)={\int}_{0}^{1}\frac{\mathrm{ln}(1-2x\mathrm{sin}a+{x}^{2})}{x}dx$

${J}^{\prime}\left(a\right)=-{\int}_{0}^{1}\frac{2\mathrm{cos}a}{{(x-\mathrm{sin}a)}^{2}+{\mathrm{cos}}^{2}a}dx=-(\frac{\pi}{2}+a)$

Then, width$J\left(0\right)={\int}_{0}^{1}\frac{\mathrm{ln}(1+{x}^{2})}{x}dx=\frac{12}{I}$

$I=-\frac{32}{J}\left(\frac{\pi}{6}\right)=-\frac{32}{J\left(0\right)+{\int}_{0}^{\frac{\pi}{6}}{J}^{\prime}\left(a\right)da}=-\frac{34}{I}+{\frac{32}{\int}}_{0}^{\frac{\pi}{6}}(\frac{\pi}{2}+a)da$

which leads to

$I=\frac{{\pi}^{2}}{12}$

Let

Then, width

which leads to

sonorous9n

Beginner2022-01-05Added 34 answers

Hint. One may recall that

$\mathrm{ln}(1+x)=\sum _{n=1}^{\mathrm{\infty}}{(-1)}^{n-1}\frac{{x}^{n}}{n},\text{}\left|x\right|1$

one may then divide by x and one is allowed to integrate termwise obtaining

$\int}_{0}^{1}\frac{\mathrm{ln}(1+x)}{x}dx=\sum _{n=1}^{\mathrm{\infty}}{(-1)}^{n-1}{\int}_{0}^{1}\frac{{x}^{n-1}}{n}=\sum _{n=1}^{\mathrm{\infty}}\frac{{(-1)}^{n-1}}{{n}^{2}$

Can you take it from here?

one may then divide by x and one is allowed to integrate termwise obtaining

Can you take it from here?

user_27qwe

Skilled2022-01-11Added 375 answers

By integration by parts:

but since

Find the local maximum and minimum values and saddle points of the function. If you have three-dimensional graphing software, graph the function with a domain and viewpoint that reveal all the important aspects of the function

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