Francisca Rodden

2022-01-03

Compute

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

scoollato7o

Beginner2022-01-04Added 26 answers

Put $x=\mathrm{tan}\theta$ , then your integral transforms to

$I={\int}_{0}^{\frac{\pi}{4}}\mathrm{log}(1+\mathrm{tan}\theta )d\theta$ (1)

Now using the property that

${\int}_{0}^{a}f\left(x\right)dx={\int}_{0}^{a}f(a-x)dx$

we have

$I={\int}_{0}^{\frac{\pi}{4}}\mathrm{log}(1+\mathrm{tan}(\frac{\pi}{4}-\theta ))d\theta ={\int}_{0}^{\frac{\pi}{4}}\mathrm{log}\left(\frac{2}{1+\mathrm{tan}\theta}\right)d\theta$ (2)

Adding (1) and (2) we get

$2I={\int}_{0}^{\frac{\pi}{4}}\mathrm{log}\left(2\right)d\theta \Rightarrow I=\mathrm{log}\left(2\right)\cdot \frac{\pi}{8}$

Now using the property that

we have

Adding (1) and (2) we get

Barbara Meeker

Beginner2022-01-05Added 38 answers

Consider:

$I\left(a\right)={\int}_{0}^{1}\frac{\mathrm{ln}(1+ax)}{1+{x}^{2}}dx$

than the derivative$I}^{\prime$ is equal:

$I}^{\prime}\left(a\right)={\int}_{0}^{1}\frac{x}{(1+ax)(1+{x}^{2})}dx=\frac{2a\mathrm{arctan}x-2\mathrm{ln}(1+ax)+\mathrm{ln}(1+{x}^{2})}{2(1+{a}^{2})}{\mid}_{0}^{1$

$=\frac{\pi a+2\mathrm{ln}2}{4(1+{a}^{2})}-\frac{\mathrm{ln}(1+a)}{1+{a}^{2}}$

Hence:

$I\left(1\right)={\int}_{0}^{1}(\frac{\pi a+2\mathrm{ln}2}{4(1+{a}^{2})}-\frac{\mathrm{ln}(1+a)}{1+{a}^{2}})da$

$2I\left(1\right)={\int}_{0}^{1}\frac{\pi a+2\mathrm{ln}2}{4(1+{a}^{2})}da=\frac{\pi}{4}\mathrm{ln}2$

Divide both sides by 2 and you're done.

than the derivative

Hence:

Divide both sides by 2 and you're done.

Vasquez

Expert2022-01-09Added 669 answers

If

and since

Therefore,

Adding the left side to both sides and dividing by 2 yields

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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