Question

x³+3x²+2x+1/x-1,Integrate the given function defined on proper domain w.r.t. x.

Answer 1

by integrating we get,
x^4/4+3x^3/3+2x^2/2+logx-1
hope u got it

Answer 2

question is -----> $\int{\frac{x^3+3x^2+2x+1}{x-1}}\,dx$

first of all,we should solve numerator (x³ + 3x² + 2x + 1) in such a way that it contains denominator (x - 1).
e.g., x³ + 3x² + 2x + 1
= x³ - x² + 4x² - 4x + 6x - 6 + 7
= x²(x - 1) + 4x(x - 1) + 6(x - 1) + 7
= (x - 1)(x² + 4x + 6) + 7

now, $\int{\frac{(x-1)(x^2+4x+6)+7}{x-1}}\,dx$

$\int{(x^2+4x+6)+\frac{7}{(x-1)}}\,dx\\\\\\=\int{x^2}\,dx+\int{4x}\,dx+6\int{dx}+\int{\frac{7}{x-1}}\,dx\\\\\\=\left[\frac{x^3}{3}\right]+4\left[\frac{x^2}{2}\right]+6x+7ln|x-1|+C\\\\\\=\frac{x^3}{3}+2x^2+6x+7ln|x-1|+C$