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Find the integral $\int\frac{x^2-9x+15}{x^3-6x^2+12x-8}dx$

Step-by-step Solution

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Final Answer

$\ln\left(x-2\right)+\frac{5}{x-2}+\frac{1}{-2\left(x-2\right)^{2}}+C_0$
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Step-by-step Solution

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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 $-8$

$1, 2, 4, 8$

Next, list all divisors of the leading coefficient $a_n$, which equals $1$

$1$

The possible roots $\pm\frac{p}{q}$ of the polynomial $x^3-6x^2+12x-8$ will then be

$\pm1,\:\pm2,\:\pm4,\:\pm8$

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

$2^3-6\cdot 2^2+12\cdot 2-8=0$

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

$\left|\begin{array}{c}1 & -6 & 12 & -8 \\ & 2 & -8 & 8 \\ 1 & -4 & 4 & 0\end{array}\right|2$

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 $\left(x-2\right)$

$\int\frac{x^2-9x+15}{\left(x^{2}-4x+4\right)\left(x-2\right)}dx$

The trinomial $\left(x^{2}-4x+4\right)$ is a perfect square trinomial, because it's discriminant is equal to zero

$\Delta=b^2-4ac=-4^2-4\left(1\right)\left(4\right) = 0$

Using the perfect square trinomial formula

$a^2+2ab+b^2=(a+b)^2,\:where\:a=\sqrt{x^{2}}\:and\:b=\sqrt{4}$

Factoring the perfect square trinomial

$\int\frac{x^2-9x+15}{\left(x-2\right)^{2}\left(x-2\right)}dx$

When multiplying exponents with same base you can add the exponents: $\left(x-2\right)^{2}\left(x-2\right)$

$\int\frac{x^2-9x+15}{\left(x-2\right)^{3}}dx$
1

Rewrite the expression $\frac{x^2-9x+15}{x^3-6x^2+12x-8}$ inside the integral in factored form

$\int\frac{x^2-9x+15}{\left(x-2\right)^{3}}dx$
2

Rewrite the fraction $\frac{x^2-9x+15}{\left(x-2\right)^{3}}$ in $3$ simpler fractions using partial fraction decomposition

$\frac{x^2-9x+15}{\left(x-2\right)^{3}}=\frac{A}{x-2}+\frac{B}{\left(x-2\right)^{2}}+\frac{C}{\left(x-2\right)^{3}}$
3

Find the values for the unknown coefficients: $A, B, C$. The first step is to multiply both sides of the equation from the previous step by $\left(x-2\right)^{3}$

$x^2-9x+15=\left(x-2\right)^{3}\left(\frac{A}{x-2}+\frac{B}{\left(x-2\right)^{2}}+\frac{C}{\left(x-2\right)^{3}}\right)$
4

Multiply both sides of the equality by $1$ to simplify the fractions

$1\left(x^2-9x+15\right)=1\left(x-2\right)^{3}\left(\frac{A}{x-2}+\frac{B}{\left(x-2\right)^{2}}+\frac{C}{\left(x-2\right)^{3}}\right)$
5

Multiplying polynomials

$1\left(x^2-9x+15\right)=\frac{1\left(x-2\right)^{3}A}{x-2}+\frac{1\left(x-2\right)^{3}B}{\left(x-2\right)^{2}}+\frac{1\left(x-2\right)^{3}C}{\left(x-2\right)^{3}}$
6

Simplifying

$1\left(x^2-9x+15\right)=\left(x-2\right)^{2}A+\left(x-2\right)B+C$
7

Assigning values to $x$ we obtain the following system of equations

$\begin{matrix}15=4A-2B+C&\:\:\:\:\:\:\:(x=0) \\ 7=A-B+C&\:\:\:\:\:\:\:(x=1) \\ 25=9A-3B+C&\:\:\:\:\:\:\:(x=-1)\end{matrix}$
8

Proceed to solve the system of linear equations

$\begin{matrix}4A & - & 2B & + & 1C & =15 \\ 1A & - & 1B & + & 1C & =7 \\ 9A & - & 3B & + & 1C & =25\end{matrix}$
9

Rewrite as a coefficient matrix

$\left(\begin{matrix}4 & -2 & 1 & 15 \\ 1 & -1 & 1 & 7 \\ 9 & -3 & 1 & 25\end{matrix}\right)$
10

Reducing the original matrix to a identity matrix using Gaussian Elimination

$\left(\begin{matrix}1 & 0 & 0 & 1 \\ 0 & 1 & 0 & -5 \\ 0 & 0 & 1 & 1\end{matrix}\right)$
11

The integral of $\frac{x^2-9x+15}{\left(x-2\right)^{3}}$ in decomposed fraction equals

$\int\left(\frac{1}{x-2}+\frac{-5}{\left(x-2\right)^{2}}+\frac{1}{\left(x-2\right)^{3}}\right)dx$
12

Expand the integral $\int\left(\frac{1}{x-2}+\frac{-5}{\left(x-2\right)^{2}}+\frac{1}{\left(x-2\right)^{3}}\right)dx$ into $3$ integrals using the sum rule for integrals, to then solve each integral separately

$\int\frac{1}{x-2}dx+\int\frac{-5}{\left(x-2\right)^{2}}dx+\int\frac{1}{\left(x-2\right)^{3}}dx$

We can solve the integral $\int\frac{1}{x-2}dx$ 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

$u=x-2$

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

$du=dx$

Substituting $u$ and $dx$ in the integral and simplify

$\int\frac{1}{u}du$

The integral of the inverse of the lineal function is given by the following formula, $\displaystyle\int\frac{1}{x}dx=\ln(x)$

$\ln\left(u\right)$

Replace $u$ with the value that we assigned to it in the beginning: $x-2$

$\ln\left(x-2\right)$
13

The integral $\int\frac{1}{x-2}dx$ results in: $\ln\left(x-2\right)$

$\ln\left(x-2\right)$

We can solve the integral $\int\frac{-5}{\left(x-2\right)^{2}}dx$ 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

$u=x-2$

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

$du=dx$

Substituting $u$ and $dx$ in the integral and simplify

$\int\frac{-5}{u^{2}}du$

Rewrite the exponent using the power rule $\frac{a^m}{a^n}=a^{m-n}$, where in this case $m=0$

$\int-5u^{-2}du$

The integral of a function times a constant ($-5$) is equal to the constant times the integral of the function

$-5\int u^{-2}du$

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 $-2$

$5u^{-1}$

Replace $u$ with the value that we assigned to it in the beginning: $x-2$

$5\left(x-2\right)^{-1}$

Applying the property of exponents, $\displaystyle a^{-n}=\frac{1}{a^n}$, where $n$ is a number

$\frac{5}{1\left(x-2\right)}$

Any expression multiplied by $1$ is equal to itself

$\frac{5}{x-2}$
14

The integral $\int\frac{-5}{\left(x-2\right)^{2}}dx$ results in: $\frac{5}{x-2}$

$\frac{5}{x-2}$

We can solve the integral $\int\frac{1}{\left(x-2\right)^{3}}dx$ 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

$u=x-2$

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

$du=dx$

Substituting $u$ and $dx$ in the integral and simplify

$\int\frac{1}{u^{3}}du$

Rewrite the exponent using the power rule $\frac{a^m}{a^n}=a^{m-n}$, where in this case $m=0$

$\int u^{-3}du$

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 $-3$

$\frac{u^{-2}}{-2}$

Applying the property of exponents, $\displaystyle a^{-n}=\frac{1}{a^n}$, where $n$ is a number

$\frac{1}{-2u^{2}}$

Replace $u$ with the value that we assigned to it in the beginning: $x-2$

$\frac{1}{-2\left(x-2\right)^{2}}$
15

The integral $\int\frac{1}{\left(x-2\right)^{3}}dx$ results in: $\frac{1}{-2\left(x-2\right)^{2}}$

$\frac{1}{-2\left(x-2\right)^{2}}$
16

Gather the results of all integrals

$\ln\left(x-2\right)+\frac{5}{x-2}+\frac{1}{-2\left(x-2\right)^{2}}$
17

As the integral that we are solving is an indefinite integral, when we finish integrating we must add the constant of integration $C$

$\ln\left(x-2\right)+\frac{5}{x-2}+\frac{1}{-2\left(x-2\right)^{2}}+C_0$

Final Answer

$\ln\left(x-2\right)+\frac{5}{x-2}+\frac{1}{-2\left(x-2\right)^{2}}+C_0$

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Function Plot

Plotting: $\ln\left(x-2\right)+\frac{5}{x-2}+\frac{1}{-2\left(x-2\right)^{2}}+C_0$

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How to improve your answer:

Main Topic: Integrals by Partial Fraction Expansion

The partial fraction decomposition or partial fraction expansion of a rational function is the operation that consists in expressing the fraction as a sum of a polynomial (possibly zero) and one or several fractions with a simpler denominator.

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