Sr Examen
Integral de (x+4)/(2x^2-7x+1) dx
Solución
Ha introducido
[src]
1 / | | x + 4 | -------------- dx | 2 | 2*x - 7*x + 1 | / 0
$$\int\limits_{0}^{1} \frac{x + 4}{\left(2 x^{2} - 7 x\right) + 1}\, dx$$
Integral((x + 4)/(2*x^2 - 7*x + 1), (x, 0, 1))
Respuesta (Indefinida)
[src]
// / ____ \ \
|| ____ |4*\/ 41 *(-7/4 + x)| |
||-\/ 41 *acoth|-------------------| |
/ || \ 41 / 2 41|
| ||----------------------------------- for (-7/4 + x) > --| / 2\
| x + 4 || 164 16| log\1 - 7*x + 2*x /
| -------------- dx = C + 46*|< | + -------------------
| 2 || / ____ \ | 4
| 2*x - 7*x + 1 || ____ |4*\/ 41 *(-7/4 + x)| |
| ||-\/ 41 *atanh|-------------------| |
/ || \ 41 / 2 41|
||----------------------------------- for (-7/4 + x) < --|
\\ 164 16/
$$\int \frac{x + 4}{\left(2 x^{2} - 7 x\right) + 1}\, dx = C + 46 \left(\begin{cases} - \frac{\sqrt{41} \operatorname{acoth}{\left(\frac{4 \sqrt{41} \left(x - \frac{7}{4}\right)}{41} \right)}}{164} & \text{for}\: \left(x - \frac{7}{4}\right)^{2} > \frac{41}{16} \\- \frac{\sqrt{41} \operatorname{atanh}{\left(\frac{4 \sqrt{41} \left(x - \frac{7}{4}\right)}{41} \right)}}{164} & \text{for}\: \left(x - \frac{7}{4}\right)^{2} < \frac{41}{16} \end{cases}\right) + \frac{\log{\left(2 x^{2} - 7 x + 1 \right)}}{4}$$
Gráfica
Estos ejemplos se pueden aplicar para introducción de los límites de integración inferior y superior.