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 .

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