Đáp án:
$ f(x)=ln\dfrac{x-1}{x+2}++5$
Giải thích các bước giải:
$f'(x)=\dfrac{3}{x^2+x-2}$
$\rightarrow \int f'(x)dx=\int \dfrac{3}{x^2+x-2} dx$
$\rightarrow f(x)=\int \dfrac{3}{(x+2)(x-1)}dx$
$\rightarrow f(x)=\int \dfrac{x+2-(x-1)}{(x+2)(x-1)}dx$
$\rightarrow f(x)=\int \dfrac{1}{x-1}-\dfrac{1}{x+2}dx$
$\rightarrow f(x)=ln(x-1)-ln(x+2)+C$
$\rightarrow f(x)=ln\dfrac{x-1}{x+2}+C$
do $f(2)=5-2ln2\rightarrow ln\dfrac{2-1}{2+2}+C=5-2ln2$
$\rightarrow ln\dfrac{1}{2^2}+C=5-2ln2$
$\rightarrow -2ln2+C=5-2ln2$
$\rightarrow C=5$
$\rightarrow f(x)=ln\dfrac{x-1}{x+2}++5$