Đáp án: $P=15$
Giải thích các bước giải:
Ta có :
$\int\dfrac{x+1}{x^2+x-6}dx$
$=\int\dfrac{x+1}{(x+3)(x-2)}dx$
$=\int\dfrac{x+3-2}{(x+3)(x-2)}dx$
$=\int\dfrac{1}{x-2}-\dfrac{2}{(x+3)(x-2)}dx$
$=\int\dfrac{1}{x-2}-\dfrac{2}{5}.\dfrac{5}{(x+3)(x-2)}dx$
$=\int\dfrac{1}{x-2}-\dfrac{2}{5}.\dfrac{x+3-(x-2)}{(x+3)(x-2)}dx$
$=\int\dfrac{1}{x-2}-\dfrac{2}{5}(\dfrac{1}{x-2}-\dfrac{1}{x+3})dx$
$=\int\dfrac{1}{x-2}-\dfrac{2}{5}.\dfrac{1}{x-2}+\dfrac25.\dfrac{1}{x+3}dx$
$=\int\dfrac{3}{5}.\dfrac{1}{x-2}+\dfrac25.\dfrac{1}{x+3}dx$
$=\dfrac35\ln|x-2|+\dfrac25\ln|x+3|$
$\to \int^3_4\dfrac{x+1}{x^2+x-6}dx=-\dfrac{3\ln \left(2\right)+2\left(\ln \left(7\right)-\ln \left(6\right)\right)}{5}$
$\to \int^3_4\dfrac{x+1}{x^2+x-6}dx=-\dfrac{3\ln \left(2\right)+2\left(\ln \left(\dfrac76\right)\right)}{5}$
$\to \int^3_4\dfrac{x+1}{x^2+x-6}dx=-\dfrac{3}{5}\ln \left(2\right)+\dfrac{2}{5}\ln \left(\dfrac76\right)$
$\to a=7,b=6,c=2\to a+b+c=15$
