Solved: "Compute ∫01ln⁡(x+1)x2+1dx"

Francisca Rodden

Answered question

2022-01-03

Compute

\int_{0}^{1} \frac{\ln (x + 1)}{x^{2} + 1} d x

Answer & Explanation

scoollato7o

Beginner2022-01-04Added 26 answers

Put

x = \tan θ

, then your integral transforms to

I = \int_{0}^{\frac{π}{4}} \log (1 + \tan θ) d θ

(1)
Now using the property that

\int_{0}^{a} f (x) d x = \int_{0}^{a} f (a - x) d x

we have

I = \int_{0}^{\frac{π}{4}} \log (1 + \tan (\frac{π}{4} - θ)) d θ = \int_{0}^{\frac{π}{4}} \log (\frac{2}{1 + \tan θ}) d θ

(2)
Adding (1) and (2) we get

2 I = \int_{0}^{\frac{π}{4}} \log (2) d θ \Rightarrow I = \log (2) \cdot \frac{π}{8}

Barbara Meeker

Beginner2022-01-05Added 38 answers

Consider:

I (a) = \int_{0}^{1} \frac{\ln (1 + a x)}{1 + x^{2}} d x

than the derivative

I^{'}

is equal:

I^{'} (a) = \int_{0}^{1} \frac{x}{(1 + a x) (1 + x^{2})} d x = \frac{2 a \arctan x - 2 \ln (1 + a x) + \ln (1 + x^{2})}{2 (1 + a^{2})} ∣_{0}^{1}

= \frac{π a + 2 \ln 2}{4 (1 + a^{2})} - \frac{\ln (1 + a)}{1 + a^{2}}

Hence:

I (1) = \int_{0}^{1} (\frac{π a + 2 \ln 2}{4 (1 + a^{2})} - \frac{\ln (1 + a)}{1 + a^{2}}) d a

2 I (1) = \int_{0}^{1} \frac{π a + 2 \ln 2}{4 (1 + a^{2})} d a = \frac{π}{4} \ln 2

Divide both sides by 2 and you're done.

Vasquez

Expert2022-01-09Added 669 answers

If $(1 + x) (1 + y) = 2$ , then
$x = \frac{1 - y}{1 + y}$
$1 + x^{2} = 2 \frac{1 + y^{2}}{(1 + y)^{2}}$
$\frac{1 + x^{2}}{1 + x} = \frac{1 + y^{2}}{1 + y}$
and since $(1 + y) d x + (1 + x) d y = 0$ we get
$\frac{d x}{1 + x^{2}} = - \frac{d y}{1 + y^{2}}$
Therefore,
$\int_{0}^{1} \frac{\log (1 + x)}{1 + x^{2}} d x = \int_{0}^{1} \frac{\log (2) - \log (1 + y)}{1 + y^{2}} d y$
Adding the left side to both sides and dividing by 2 yields
$\int_{0}^{1} \frac{\log (1 + x)}{1 + x^{2}} d x = \frac{1}{2} \int_{0}^{1} \frac{\log (2)}{1 + y^{2}} d y$
$= \frac{π}{8} \log (2)$

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