Интеграл...................................

Решите задачу:

$\int\ {\frac{x^3-1}{x+1} \, dx = \int\ {(\frac{x^3}{x+1} - \frac{1}{x+1}) } \, dx = \int\ {\frac{x^3}{x+1} \, dx - \int \frac{1}{x+1} }} \, dx= \\ \\ = \int\ {\frac{x^2 \cdot (x+1) -x \cdot (x+1)+(x+1)-1}{x+1} \, dx - \int \frac{1}{x+1} } \, dx= \\ \\ = \int\ {(x^2 -x +1-\frac{1}{x+1}) \, dx - \ln{|x+1|}} =\\ \\ = \int\ {x^2} \, dx - \int\ {x } \, dx + \int\ {1} \, dx - \int\ {\frac{1}{x+1}} \, dx- \ln{|x+1|}= \\ \\ = \frac{x^3}{3}-\frac{x^2}{2}+x-\ln{|x+1|}-\ln{|x+1|}+C=$ $\\ \\ = \frac{x^3}{3}-\frac{x^2}{2}+x-2\ln{|x+1|}+C=$