Question

Решить определённые интегралы (пошагово)

Answer 1

1. $\int\limits^1_0 {\frac{x}{x^2-5x+6} } \, dx = \frac{1}{2} \int\limits^1_0 {\frac{1}{x^2-5x+6} } \, dx^2 =  \frac{1}{2} \int\limits^1_0 {\frac{1}{x^2-5x+\frac{25}{4}+6-\frac{25}{4}  } } \, dx^2 = \frac{1}{2} \int\limits^1_0 {\frac{1}{(x-\frac{5}{2})^2-\frac{1}{4}  } } \, dx^2 = \frac{1}{2} ln|(x-\frac{5}{2})^2-\frac{1}{4}  | |\frac{1}{0} = \frac{1}{2} (ln\frac{9}{4} -\frac{1}{4} -ln\frac{25}{4} +\frac{1}{4} ) = \frac{1}{2} (ln\frac{9}{4} - ln\frac{25}{4} ) = \frac{1}{2} ln\frac{9*4}{4*25} =$ $\frac{1}{2} ln\frac{9}{25}$

2. $\int\limits^e_1 {x^2log(x)} \, dx = |u=log(x), dv = x^2dx, du = \frac{1}{xlna} dx, v = \frac{x^3}{3} | = \frac{log(x)x^3}{3} - \int\limits^e_1 {\frac{x^3}{3xlna} } \, dx = \frac{log(x)x^3}{3} - \frac{1}{3lna} \int\limits^e_1 {x^2 } \, dx = \frac{log(x)x^3}{3} - \frac{1}{3lna} *\frac{x^3}{3} |\frac{e}{1} = \frac{log(e)e^3}{3} - \frac{1}{3lna} * \frac{e^3}{3} + \frac{1}{3lna} *\frac{1}{3} = \frac{log(e)e^3}{3} - \frac{1}{3lna} (\frac{e^3-1}{3} )$