Giải thích các bước giải:
$a.\int\dfrac{x^2-2x+3}{x+1}dx$
$=\int\dfrac{(x+1)^2-4(x+1)+6}{x+1}dx$
$=\int x+1-4+\dfrac{6}{x+1}dx$
$=\int x-3+\dfrac{6}{x+1}dx$
$=\dfrac{x^2}{2}-3x+6\ln|x+1|+C $
b.$\int\dfrac{x^3}{x-1}dx$
$=\int\dfrac{x^3-1+1}{x-1}dx$
$=\int\dfrac{(x-1)(x^2+x+1)+1}{x-1}dx$
$=\int x^2+x+1+\dfrac{1}{x-1}dx$
$=\dfrac{x^3}{3}+\dfrac{x^2}{2}+x+\ln |x-1|+C$