Sr Examen
Integral de (x-1)/(x+1)lnx dx
Solución
Ha introducido
[src]
1 / | | x - 1 | -----*log(x) dx | x + 1 | / 0
$$\int\limits_{0}^{1} \frac{x - 1}{x + 1} \log{\left(x \right)}\, dx$$
Integral(((x - 1)/(x + 1))*log(x), (x, 0, 1))
Respuesta (Indefinida)
[src]
// -polylog(2, 1 + x) + pi*I*log(1 + x) for |1 + x| < 1\
/ || |
| || / 1 \ 1 |
| x - 1 || -polylog(2, 1 + x) - pi*I*log|-----| for ------- < 1| / pi*I\
| -----*log(x) dx = C - x - |< \1 + x/ |1 + x| | - polylog\2, x*e / + (x - log(1 + x))*log(x)
| x + 1 || |
| || __0, 2 /1, 1 | \ __2, 0 / 1, 1 | \ |
/ ||-polylog(2, 1 + x) + pi*I*/__ | | 1 + x| - pi*I*/__ | | 1 + x| otherwise |
\\ \_|2, 2 \ 0, 0 | / \_|2, 2 \0, 0 | / /
$$\int \frac{x - 1}{x + 1} \log{\left(x \right)}\, dx = C - x + \left(x - \log{\left(x + 1 \right)}\right) \log{\left(x \right)} - \begin{cases} i \pi \log{\left(x + 1 \right)} - \operatorname{Li}_{2}\left(x + 1\right) & \text{for}\: \left|{x + 1}\right| < 1 \\- i \pi \log{\left(\frac{1}{x + 1} \right)} - \operatorname{Li}_{2}\left(x + 1\right) & \text{for}\: \frac{1}{\left|{x + 1}\right|} < 1 \\- i \pi {G_{2, 2}^{2, 0}\left(\begin{matrix} & 1, 1 \\0, 0 & \end{matrix} \middle| {x + 1} \right)} + i \pi {G_{2, 2}^{0, 2}\left(\begin{matrix} 1, 1 & \\ & 0, 0 \end{matrix} \middle| {x + 1} \right)} - \operatorname{Li}_{2}\left(x + 1\right) & \text{otherwise} \end{cases} - \operatorname{Li}_{2}\left(x e^{i \pi}\right)$$
Respuesta
[src]
2
pi
-1 + --- - 2*pi*I*log(2)
6
$$-1 + \frac{\pi^{2}}{6} - 2 i \pi \log{\left(2 \right)}$$
=
=
2
pi
-1 + --- - 2*pi*I*log(2)
6
$$-1 + \frac{\pi^{2}}{6} - 2 i \pi \log{\left(2 \right)}$$
-1 + pi^2/6 - 2*pi*i*log(2)
Estos ejemplos se pueden aplicar para introducción de los límites de integración inferior y superior.