Sr Examen
Integral de logx/(1+x) dx
Solución
Ha introducido
[src]
1 / | | log(x) | ------ dx | 1 + x | / 1/10
$$\int\limits_{\frac{1}{10}}^{1} \frac{\log{\left(x \right)}}{x + 1}\, dx$$
Integral(log(x)/(1 + x), (x, 1/10, 1))
Respuesta (Indefinida)
[src]
// -polylog(2, 1 + x) + pi*I*log(1 + x) for |1 + x| < 1\
/ || |
| || / 1 \ 1 |
| log(x) || -polylog(2, 1 + x) - pi*I*log|-----| for ------- < 1|
| ------ dx = C + |< \1 + x/ |1 + x| |
| 1 + x || |
| || __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 | / /
$$\int \frac{\log{\left(x \right)}}{x + 1}\, dx = C + \begin{cases} i \pi \log{\left(x + 1 \right)} - \operatorname{Li}_{2}\left(x + 1\right) & \text{for}\: \left|{x + 1}\right| < 1 \\- i \pi \log{\left(\frac{1}{x + 1} \right)} - \operatorname{Li}_{2}\left(x + 1\right) & \text{for}\: \frac{1}{\left|{x + 1}\right|} < 1 \\- i \pi {G_{2, 2}^{2, 0}\left(\begin{matrix} & 1, 1 \\0, 0 & \end{matrix} \middle| {x + 1} \right)} + i \pi {G_{2, 2}^{0, 2}\left(\begin{matrix} 1, 1 & \\ & 0, 0 \end{matrix} \middle| {x + 1} \right)} - \operatorname{Li}_{2}\left(x + 1\right) & \text{otherwise} \end{cases}$$
Gráfica
Respuesta
[src]
2 pi /11\ / 11\ - --- - pi*I*log|--| + 2*pi*I*log(2) + polylog|2, --| 4 \10/ \ 10/
$$- \frac{\pi^{2}}{4} - i \pi \log{\left(\frac{11}{10} \right)} + \operatorname{Li}_{2}\left(\frac{11}{10}\right) + 2 i \pi \log{\left(2 \right)}$$
=
=
2 pi /11\ / 11\ - --- - pi*I*log|--| + 2*pi*I*log(2) + polylog|2, --| 4 \10/ \ 10/
$$- \frac{\pi^{2}}{4} - i \pi \log{\left(\frac{11}{10} \right)} + \operatorname{Li}_{2}\left(\frac{11}{10}\right) + 2 i \pi \log{\left(2 \right)}$$
-pi^2/4 - pi*i*log(11/10) + 2*pi*i*log(2) + polylog(2, 11/10)
Estos ejemplos se pueden aplicar para introducción de los límites de integración inferior y superior.