Francisca Rodden

## Answered question

2022-01-03

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

### Answer & Explanation

scoollato7o

Beginner2022-01-04Added 26 answers

Put $x=\mathrm{tan}\theta$, then your integral transforms to
$I={\int }_{0}^{\frac{\pi }{4}}\mathrm{log}\left(1+\mathrm{tan}\theta \right)d\theta$ (1)
Now using the property that
${\int }_{0}^{a}f\left(x\right)dx={\int }_{0}^{a}f\left(a-x\right)dx$
we have
$I={\int }_{0}^{\frac{\pi }{4}}\mathrm{log}\left(1+\mathrm{tan}\left(\frac{\pi }{4}-\theta \right)\right)d\theta ={\int }_{0}^{\frac{\pi }{4}}\mathrm{log}\left(\frac{2}{1+\mathrm{tan}\theta }\right)d\theta$ (2)
Adding (1) and (2) we get
$2I={\int }_{0}^{\frac{\pi }{4}}\mathrm{log}\left(2\right)d\theta ⇒I=\mathrm{log}\left(2\right)\cdot \frac{\pi }{8}$

Barbara Meeker

Beginner2022-01-05Added 38 answers

Consider:
$I\left(a\right)={\int }_{0}^{1}\frac{\mathrm{ln}\left(1+ax\right)}{1+{x}^{2}}dx$
than the derivative ${I}^{\prime }$ is equal:
${I}^{\prime }\left(a\right)={\int }_{0}^{1}\frac{x}{\left(1+ax\right)\left(1+{x}^{2}\right)}dx=\frac{2a\mathrm{arctan}x-2\mathrm{ln}\left(1+ax\right)+\mathrm{ln}\left(1+{x}^{2}\right)}{2\left(1+{a}^{2}\right)}{\mid }_{0}^{1}$
$=\frac{\pi a+2\mathrm{ln}2}{4\left(1+{a}^{2}\right)}-\frac{\mathrm{ln}\left(1+a\right)}{1+{a}^{2}}$
Hence:
$I\left(1\right)={\int }_{0}^{1}\left(\frac{\pi a+2\mathrm{ln}2}{4\left(1+{a}^{2}\right)}-\frac{\mathrm{ln}\left(1+a\right)}{1+{a}^{2}}\right)da$
$2I\left(1\right)={\int }_{0}^{1}\frac{\pi a+2\mathrm{ln}2}{4\left(1+{a}^{2}\right)}da=\frac{\pi }{4}\mathrm{ln}2$
Divide both sides by 2 and you're done.

Vasquez

Expert2022-01-09Added 669 answers

If $\left(1+x\right)\left(1+y\right)=2$, then
$x=\frac{1-y}{1+y}$
$1+{x}^{2}=2\frac{1+{y}^{2}}{\left(1+y{\right)}^{2}}$
$\frac{1+{x}^{2}}{1+x}=\frac{1+{y}^{2}}{1+y}$
and since $\left(1+y\right)dx+\left(1+x\right)dy=0$ we get
$\frac{dx}{1+{x}^{2}}=-\frac{dy}{1+{y}^{2}}$
Therefore,
${\int }_{0}^{1}\frac{\mathrm{log}\left(1+x\right)}{1+{x}^{2}}dx={\int }_{0}^{1}\frac{\mathrm{log}\left(2\right)-\mathrm{log}\left(1+y\right)}{1+{y}^{2}}dy$
Adding the left side to both sides and dividing by 2 yields
${\int }_{0}^{1}\frac{\mathrm{log}\left(1+x\right)}{1+{x}^{2}}dx=\frac{1}{2}{\int }_{0}^{1}\frac{\mathrm{log}\left(2\right)}{1+{y}^{2}}dy$
$=\frac{\pi }{8}\mathrm{log}\left(2\right)$

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