Indefinite integral of a rational function with linear denominator: $ \int \frac{ x^7}{(x+1)}{dx} $

Question

$$ \int \frac{ x^7}{(x+1)}{dx} $$ $$ \int \frac{ \left(x^7 + x^6 - x^6 - x^5 + x^5 + x^4 -x^4 - x^3 + x^3 + x^2 - x^2 -x^1 + x^1 +1 -1\right ) }{\left(x+1\right)}{dx}$$ $$ \int { \left(x^6 - x^5 + x^4 - x^3 + x^2 -x +1 - \frac{1}{x+1}\right)}{dx}$$

$$\frac{x^7}{7} - \frac{x^6}{6} + \frac{x^5}{5} -\frac{x^4}{4}+ \frac{x^3}{3} - \frac{x^2}{2} + \frac{x^1}{1} -\frac{\log(x+1)}{1} +C $$ Is there any other good way to do this ?

Answer 1

Let $$I=\int\frac{x^7}{x+1}dx$$ Apply $x+1\to y$ to get $$I=\int\frac{(y-1)^7}{y}dy$$ Using the binomial theorem $$I=\int\frac{1}{y}\sum_{k=0}^{7}\binom{7}{k}y^k(-1)^{7-k}dy$$ $$=\int\sum_{k=0}^{7}\binom{7}{k}y^{k-1}(-1)^{7-k}dy$$ $$=\sum_{k=0}^{7}\int\binom{7}{k}y^{k-1}(-1)^{7-k}dy$$ $$=C-\ln|y|+\sum_{k=1}^{7}\frac{1}{k}\binom{7}{k}y^k(-1)^{7-k}$$ $$=C-\ln|x+1|+\sum_{k=1}^{7}\frac{1}{k}\binom{7}{k}(x+1)^k(-1)^{7-k}$$ At which point you have by most standards a satisfactory answer.

EDIT
If you teacher did some geometric progression stuff this might have been what he did: $$I=\int\frac{x^7}{x+1}dx$$ $$=\int\frac{y^7}{1-y}dy\qquad(x\to -y)$$ $$=-\int\frac{-1+1-y^7}{1-y}dy$$ $$=\int\frac{1}{1-y}dy-\int\frac{1-y^7}{1-y}dy$$ $$=\ln|1-y|-\int\left(\sum_{n=0}^6y^n\right)dy$$ $$=\ln|1-y|-\sum_{n=0}^6\int y^ndy$$ $$=\ln|1-y|-\sum_{n=1}^7\frac{y^n}{n}$$ $$=\ln|1+x|-\sum_{n=1}^7\frac{(-x)^n}{n}$$ That method there makes use of the partial sum formula for a geometric series.

Answer 2

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 $I_7$