Giải thích các bước giải:
a.Ta có:
$\displaystyle\int(\dfrac{3x+2}{x-1})^2dx$
$=\displaystyle\int(\dfrac{3(x-1)+5}{x-1})^2dx$
$=\displaystyle\int(3+\dfrac{5}{x-1})^2dx$
$=\displaystyle\int 9+\dfrac{30}{x-1}+\dfrac{25}{(x-1)^2}dx$
$=9x+30\ln|x-1|-\dfrac{25}{x-1}+C$
b.Ta có:
$\displaystyle\int \dfrac{\cos x}{3\sin^2x+\sin x-4}dx$
$=\displaystyle\int \dfrac{1}{3\sin^2x+\sin x-4}d(\sin x)$
$=\displaystyle\int \dfrac{1}{(3\sin x+4)(\sin x-1)}d(\sin x)$
$=\dfrac17\displaystyle\int \dfrac{7}{(3\sin x+4)(\sin x-1)}d(\sin x)$
$=\dfrac17\displaystyle\int \dfrac{(3\sin x+4)-3(\sin x-1)}{(3\sin x+4)(\sin x-1)}d(\sin x)$
$=\dfrac17\displaystyle\int \dfrac{1}{\sin x-1}-\dfrac{3}{(3\sin x+4)(\sin x-1)}d(\sin x)$
$=\dfrac17(\ln|\sin x-1|-\ln|3\sin x+4|)+C$
