0-1-x-ln-2-x-1-x-2-dx-

Question Number 216754 by Tawa11 last updated on 17/Feb/25

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

Answered by issac last updated on 18/Feb/25

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

u = atan (x)

\frac{d u}{d x} = \frac{1}{x^{2} + 1} \to d u = \frac{d x}{x^{2} + 1}

\int_{0}^{\frac{π}{4}} \tan (u) \ln^{2} (\tan (u)) d u \dots

so Hard to me \dots .

wolfram {alpha}^{'} s answer is

\int_{0}^{1} \frac{x \ln^{2} (x)}{x^{2} + 1} d x = \frac{3}{16} ζ (3)

ζ (s) = \underset{h = 1}{\sum^{\infty}} \frac{1}{h^{s}}

Commented by Frix last updated on 19/Feb/25

If you cannot solve it, {don}^{'} t solve it . {It}^{'} s

really that easy . {WolframAlpha}^{'} s answer

is useless, we need the path .

Commented by mathmax last updated on 23/Feb/25

3 / 16 ξ (3) i s n o t c o r r e c t t h e a n s w e r i s

l n 2 - \frac{π}{2} + 2 k_{o} k_{0} i s c a t a l a n c o n s t a n t e

Commented by Tawa11 last updated on 23/Feb/25

Please workings sir .

Answered by Frix last updated on 19/Feb/25

Some steps :

1

\int \frac{x \ln^{2} x}{x^{2} + 1} d x = \frac{1}{2} \int \frac{\ln^{2} x}{x - i} d x + \frac{1}{2} \int \frac{\ln^{2} x}{x + i} d x

2

\int \frac{\ln^{2} x}{x + a} d x \overset{[t = x + a]}{=} \int \frac{\ln^{2} (t - a)}{t} d t \overset{[by parts]}{=}

= \ln \frac{t}{a} \ln^{2} (t - a) - 2 \int \frac{\ln \frac{t}{a} \ln (t - a)}{t - a} d t

3

\int \frac{\ln \frac{t}{a} \ln (t - a)}{t - a} d t \overset{[by parts]}{=}

= - \ln (t - a) {Li}_{2} (1 - \frac{t}{a}) + \int \frac{{Li}_{2} (1 - \frac{t}{a})}{t - a} d t

4

\int \frac{{Li}_{2} (1 - \frac{t}{a})}{t - a} d t \overset{[u = 1 - \frac{t}{a}]}{=} \int \frac{{Li}_{2} u}{u} d u = {Li}_{3} u

Not sure what you learned and how {you}^{'} re

supposed to get the answer \dots

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