$\int\limits {\frac{x^{3}+2x+1}{x^{2}+6x+15}} \, dx$

Đáp án:

$\dfrac{23}{2}\ln(x^2 + 6x +15) + {22}{\sqrt6}\arctan\left(\dfrac{x+3}{\sqrt6}\right) + \dfrac{x^2}{2} - 6x + C$

Giải thích các bước giải:

$\quad I = \displaystyle\int\dfrac{x^3 + 2x +1}{x^2 + 6x +15}dx$

$\to I = \displaystyle\int\left[\dfrac{23(2x +6)}{2(x^2 + 6x +15)} + \dfrac{22}{x^2 + 6x +15} + x -6\right]dx$

$\to I = \dfrac{23}{2}\displaystyle\int\dfrac{2x+6}{x^2 + 6x +15}dx + 22\displaystyle\int\dfrac{1}{x^2 + 6x +15}dx + \displaystyle\int xdx - 6\displaystyle\int dx$

$\to I = \dfrac{23}{2}\displaystyle\int\dfrac{d(x^2 + 6x +15)}{x^2 + 6x +15}+ 22\displaystyle\int\dfrac{1}{(x+3)^2+ 6}dx + \displaystyle\int xdx - 6\displaystyle\int dx$

$\to I = \dfrac{23}{2}\ln(x^2 + 6x +15) + {22}{\sqrt6}\arctan\left(\dfrac{x+3}{\sqrt6}\right) + \dfrac{x^2}{2} - 6x + C$