Nguyên hàm của x/(x+1)^7

Đáp án: $\frac{-1}{5}$.$(x+1)^{-5}$ + $\frac{1}{6}$.$(x+1)^{-6}$ + C

Giải thích các bước giải:

$\int\limits{\frac{x}{(x+1)^7}} \, dx$ 

= $\int\limits{\frac{(x+1)-1}{(x+1)^7}} \, dx$ 

= $\int\limits{(\frac{1}{(x+1)^6}-\frac{1}{(x+1)^7})} \, dx$ 

= $\int\limits{\frac{1}{(x+1)^6}} \, dx$ - $\int\limits{\frac{1}{(x+1)^7}} \, dx$ 

= $\int\limits{\frac{d(x+1)}{(x+1)^6}} \, $ - $\int\limits{\frac{d(x+1)}{(x+1)^7}} \, $ 

= $\int\limits{{(x+1)^{-6}}} \, d(x+1)$ - $\int\limits{{(x+1)^{-7}}} \, d(x+1)$

= $\frac{-1}{5}$.$(x+1)^{-5}$ - $\frac{-1}{6}$.$(x+1)^{-6}$ + C

= $\frac{-1}{5}$.$(x+1)^{-5}$ + $\frac{1}{6}$.$(x+1)^{-6}$ + C