int (x^3 - 3x^2 + 5)/(x - 3) dx Find the indefinite integral.

int (x^3-3x^2+5)/(x-3)dx
To solve, divide the numerator by the denominator.
= int (x^2 + 5/(x-3))dx
Express it as sum of two integrals.
= int x^2dx + int 5/(x-3)dx
For the first integral, apply the formula int x^n dx = x^(n+1)/(n+1) + C .
= x^3/3 + C + int 5/(x-3)dx
For the second integral, apply u-substitution method.
Let 
u = x-3
Differentiate the u.
du = dx
Then, plug-in them to the second integral.
=x^3/3+C +5 int 1/(x-3)dx
=x^3/3+C + 5int1/udu
Apply the integral formula int 1/xdx = ln|x| + C .
= x^3/3 + 5ln|u| + C
And, substitute back u = x - 3 .
= x^3/3+ 5ln|x-3| + C
 
Therefore, int (x^3-3x^2+5)/(x-3)dx = x^3/3+5ln|x-3|+C .