Final Answer
Step-by-step Solution
Specify the solving method
We can factor the polynomial $x^3-6x^2+12x-8$ using the rational root theorem, which guarantees that for a polynomial of the form $a_nx^n+a_{n-1}x^{n-1}+\dots+a_0$ there is a rational root of the form $\pm\frac{p}{q}$, where $p$ belongs to the divisors of the constant term $a_0$, and $q$ belongs to the divisors of the leading coefficient $a_n$. List all divisors $p$ of the constant term $a_0$, which equals $-8$
Next, list all divisors of the leading coefficient $a_n$, which equals $1$
The possible roots $\pm\frac{p}{q}$ of the polynomial $x^3-6x^2+12x-8$ will then be
Trying all possible roots, we found that $2$ is a root of the polynomial. When we evaluate it in the polynomial, it gives us $0$ as a result
Now, divide the polynomial by the root we found $\left(x-2\right)$ using synthetic division (Ruffini's rule). First, write the coefficients of the terms of the numerator in descending order. Then, take the first coefficient $1$ and multiply by the factor $2$. Add the result to the second coefficient and then multiply this by $2$ and so on
In the last row of the division appear the new coefficients, with remainder equals zero. Now, rewrite the polynomial (a degree less) with the new coefficients, and multiplied by the factor $\left(x-2\right)$
The trinomial $\left(x^{2}-4x+4\right)$ is a perfect square trinomial, because it's discriminant is equal to zero
Using the perfect square trinomial formula
Factoring the perfect square trinomial
When multiplying exponents with same base you can add the exponents: $\left(x-2\right)^{2}\left(x-2\right)$
We can solve the integral $\int\frac{x^2-9x+15}{\left(x-2\right)^{3}}dx$ by applying integration by substitution method (also called U-Substitution). First, we must identify a section within the integral with a new variable (let's call it $u$), which when substituted makes the integral easier. We see that $x-2$ it's a good candidate for substitution. Let's define a variable $u$ and assign it to the choosen part
Now, in order to rewrite $dx$ in terms of $du$, we need to find the derivative of $u$. We need to calculate $du$, we can do that by deriving the equation above
Rewriting $x$ in terms of $u$
Substituting $u$, $dx$ and $x$ in the integral and simplify
Expand the fraction $\frac{-3+\left(u+2\right)^2-9u}{u^{3}}$ into $3$ simpler fractions with common denominator $u^{3}$
Simplify the resulting fractions
Expand the integral $\int\left(\frac{-3}{u^{3}}+\frac{\left(u+2\right)^2}{u^{3}}+\frac{-9}{u^{2}}\right)du$ into $3$ integrals using the sum rule for integrals, to then solve each integral separately
Rewrite the fraction $\frac{\left(u+2\right)^2}{u^{3}}$ inside the integral as the product of two functions: $\left(u+2\right)^2\frac{1}{u^{3}}$
We can solve the integral $\int\left(u+2\right)^2\frac{1}{u^{3}}du$ by applying integration by parts method to calculate the integral of the product of two functions, using the following formula
First, identify $u$ and calculate $du$
Now, identify $dv$ and calculate $v$
Solve the integral
Rewrite the exponent using the power rule $\frac{a^m}{a^n}=a^{m-n}$, where in this case $m=0$
Apply the power rule for integration, $\displaystyle\int x^n dx=\frac{x^{n+1}}{n+1}$, where $n$ represents a number or constant function, such as $-3$
Applying the property of exponents, $\displaystyle a^{-n}=\frac{1}{a^n}$, where $n$ is a number
Now replace the values of $u$, $du$ and $v$ in the last formula
Multiplying the fraction by $u+2$
Replace $u$ with the value that we assigned to it in the beginning: $x-2$
The integral $\int\frac{-3}{u^{3}}du$ results in: $\frac{3}{2\left(x-2\right)^{2}}$
The integral $-2\int\frac{u+2}{-2u^{2}}du$ results in: $\ln\left(x-2\right)+\frac{-2}{x-2}$
Gather the results of all integrals
The integral $\int\frac{-9}{u^{2}}du$ results in: $\frac{9}{x-2}$
Gather the results of all integrals
As the integral that we are solving is an indefinite integral, when we finish integrating we must add the constant of integration $C$