Tìm nguyên hàm I= Nguyên hàm của ((x^2002)/(x+2)^2005)dx

* Hướng dẫn giải

Đáp án D

$I=\int \frac{x^{2002}}{{\left(x+2\right)}^{2005}}\text{\hspace{0.17em}d}x=\int {\left(\frac{x}{x+2}\right)}^{2002}.\frac{1}{x+2}.\frac{1}{{\left(x+2\right)}^2}\text{\hspace{0.17em}\hspace{0.17em}d}x$

Đặt $t=\frac{x}{x+2}\Rightarrow \text{d}t=\frac{2.\text{d}x}{{\left(x+2\right)}^2}$

$I=\int t^{2002}.\frac{1-t}{2}.\frac{\text{d}t}{2}=\frac{1}{4}\int \left(t^{2002}-t^{2003}\right).\text{\hspace{0.17em}d}t=\frac{1}{8012}.t^{2003}-\frac{1}{8016}.t^{2004}+C$

$\Rightarrow I=\frac{1}{8012}.{\left(\frac{x}{x+2}\right)}^{2003}-\frac{1}{8016}.{\left(\frac{x}{x+2}\right)}^{2004}+C$