To perform this integration, we can first split the function into two parts to integrate each part separately.
The original integral \[ \int \frac{x^{3}-4 x^{2}+1}{x-4} d x\] separates into two integrals - one integral for the polynomial part and one for the fraction part.
The polynomial integral and its solution:
\[\int (x^{3}-4 x^{2}) dx = \frac{x^{4}}{4}- \frac{4x^{3}}{3} + C_1\]
Note that the integral of a power of x is given by \[\frac{x^{n+1}}{n+1}\]
The second integral deals with a simple fraction, x and 4 are in numerator and denominator respectively:
\[\int ( \frac{1}{x-4}) dx\]
This equation characterizes a logarithmic function, where the base, a, raised to the output, y, gives the input, x.
The integral of 1/(x-a) with respect to x is:
\[ln|x- a| + C_2\]
Because we are using absolute value, we will have logarithmic function |x − a|. Here, a is 4.
So, the integral of the fraction \[\frac{1}{x-4}\] is \[ln|x- 4|\]
Now, we add the results of the two integrals, and hence, our final solution is:
\[\frac{x^{4}}{4}- \frac{4x^{3}}{3} + ln|x- 4| + C\]
Where C is the constant of integration, and is the sum of \(C1\) and \(C2\).
So, you can see that the integral of \[\frac{x^{3}-4 x^{2}+1}{x-4}\] with respect to x is \[\frac{x^{4}}{4}- \frac{4x^{3}}{3} + ln|x- 4| + C\]