Question

Интеграл (x^2+3x-4)*lnx dx= решите подробно. Не чисто ответ!

Answer 1

$\displaystyle \int u \: \mathrm dv = uv - \int v \: \mathrm du$

$u = \ln x \\ \mathrm dv = x^2 + 3x - 4, \quad v = \dfrac{x^3}{3} + \dfrac{3x^2}{2} - 4x$

$\displaystyle \int (x^2 + 3x - 4) \ln x \: \mathrm dx = \int u \: \mathrm dv = \\ \\ \\ = \ln x \left( \dfrac{x^3}{3} + \dfrac{3x^2}{2} - 4x\right) - \int \left( \dfrac{x^3}{3} + \dfrac{3x^2}{2} - 4x\right) \dfrac{\mathrm dx}{x} = \\ \\ \\ = \ln x \left( \dfrac{x^3}{3} + \dfrac{3x^2}{2} - 4x\right) - \int \left( \dfrac{x^2}{3} + \dfrac{3x^}{2} - 4\right) \: \mathrm dx = \\ \\ \\ = \ln x \left( \dfrac{x^3}{3} + \dfrac{3x^2}{2} - 4x\right) - \left( \dfrac{x^3}{9} + \dfrac{3x^2}{4} - 4x\right) + \mathrm{const}$