Question
Find the integral: $\int \frac{x^{3}-x^{2}+x-1}{x-1} d x$

Answer

I = $\int \frac{x^{3}-x^{2}+x-1}{x-1} d x$ 
Now the numerator can be factorized as,
x3 - x2 + x - 1 = x2(x - 1) + 1(x - 1)
x3 - x2 + x - 1 = (x2 + 1)(x - 1)
Now putting this in given integral we get,
I = $\frac{x^{3}-x^{2}+x-1}{x-1}=\frac{\left(x^{2}+1\right)(x-1)}{x-1}=x^{2}+1$ 
= $\int\left(x^{2}+1\right) d x$ 
= $\int x^{2} d x+\int 1 . d x$ 
= $\frac{x^{3}}{3}+x+C$

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