Question

Integrate:

$\int \frac{{(x-1)}^2}{x^3-3x^2+3x+8}dx$

Answer 1

Consider the given integral.

$I=\int \frac{{(x-1)}^2}{x^3-3x^2+3x+8}dx$

Let $t=x^3-3x^2+3x+8$

$\frac{dt}{dx}=3x^2-6x+3$

$\frac{dt}{3}=x^2-2x+1$

$\frac{dt}{3}={(x-1)}^2$

Therefore,

$I=\frac{1}{3}\int \frac{1}{t}dt$

$I=\frac{1}{3}\ln \left(t\right)+C$

On putting the value of $t$, we get

$I=\frac{1}{3}\ln (x^3-3x^2+3x+8)+C$

Hence, this is the answer.