tìm nguyên hàm của a) (3x+2/x-1)^2dx b) cosx/3sin^2x+sinx-4 dx

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:

$\displaystyle\int \dfrac{\cos x}{3\sin^2x+\sin x-4}dx$

$=\displaystyle\int \dfrac{1}{3\sin^2x+\sin x-4}d(\sin x)$

$=\displaystyle\int \dfrac{1}{(3\sin x+4)(\sin x-1)}d(\sin x)$

$=\dfrac17\displaystyle\int \dfrac{7}{(3\sin x+4)(\sin x-1)}d(\sin x)$

$=\dfrac17\displaystyle\int \dfrac{(3\sin x+4)-3(\sin x-1)}{(3\sin x+4)(\sin x-1)}d(\sin x)$

$=\dfrac17\displaystyle\int \dfrac{1}{\sin x-1}-\dfrac{3}{(3\sin x+4)(\sin x-1)}d(\sin x)$

$=\dfrac17(\ln|\sin x-1|-\ln|3\sin x+4|)+C$