tính nguyên hàm sau: (-x^2+1)dx/x+2 Giúp Mình với. Cảm ơn nhìu.

Đặt $t=x+2$

$\to dt=dx$

Có $t=x+2\to x=t-2\to x^2=(t-2)^2$

$I=\displaystyle\int \dfrac{1-(t-2)^2}{t}dt$

$=\displaystyle\int \dfrac{1-(t^2-4t+4)}{t}dt$

$=\displaystyle\int \dfrac{-t^2+4t-3}{t}dt$

$=\displaystyle\int \left( -t+4-\dfrac{3}{t}\right)dt$

$=\dfrac{-1}{2}t^2+4t-3\ln |t|+C$

$=\dfrac{-1}{2}(x+2)^2+4(x+2)-3\ln|x+2|+C$

$=\dfrac{-1}{2}x^2+2x-3\ln|x+2|+C$

Đáp án:

$\begin{array}{l}
\int {\frac{{ - {x^2} + 1}}{{x + 2}}dx} \\
 = \int {\frac{{ - {x^2} + 4 - 3}}{{x + 2}}dx} \\
 = \int {\left( { - x - 2 - \frac{3}{{x + 2}}} \right)dx} \\
 =  - \frac{{{x^2}}}{2} - 2x - 3\ln \left| {x + 2} \right| + C
\end{array}$