<s>[INST] Indefinite integral of a rational function with linear denominator: $ \int \frac{ x^7}{(x+1)}{dx} $ $$ \int \frac{ x^7}{(x+1)}{dx} $$
$$ \int \frac{ \left(x^7 + x^6 - x^6 - x^5 + x^5 + x^4 -x^4 - x^3 + x^3 + x^2 - x^2 -x^1 + x^1 +1 -1\right ) }{\left(x+1\right)}{dx}$$
$$ \int { \left(x^6 -  x^5 + x^4  - x^3  + x^2 -x +1 - \frac{1}{x+1}\right)}{dx}$$
$$\frac{x^7}{7} - \frac{x^6}{6} + \frac{x^5}{5} -\frac{x^4}{4}+ \frac{x^3}{3} - \frac{x^2}{2} + \frac{x^1}{1} -\frac{\log(x+1)}{1}  +C $$
Is there any other good way to do this  ?
 [/INST] Same as your method : But try to use reduction formulas for higher degrees :
Let
$$I_n= \int \frac{x^n}{(x+a)}dx$$
$$I_n= \int \frac{(x+a)x^{n-1}-ax^{n-1}}{(x+a)}dx$$
$$I_n= \int x^{n-1}dx-a\int \frac{x^{n-1}}{(x+a)}dx$$
So $$I_n+aI_{n-1}=\frac{x^n}{n}$$
Now you have to find $I_7$
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