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Find the integral $\int\frac{x+8}{x^6-2x^4+x^2}dx$

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Final Answer

$\frac{-9}{4\left(x-1\right)}+\frac{-7}{4\left(x+1\right)}+\frac{-32}{37x}-2.943243\ln\left(x-1\right)+2.714414\ln\left(x+1\right)+\frac{49}{370}\ln\left(x\right)+C_0$
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Step-by-step Solution

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We can factor the polynomial $x^6-2x^4+x^2$ 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 $0$

$1$

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

$1$

The possible roots $\pm\frac{p}{q}$ of the polynomial $x^6-2x^4+x^2$ will then be

$\pm5,\:\pm4,\:\pm3,\:\pm2,\:\pm1$

We can factor the polynomial $x^6-2x^4+x^2$ using synthetic division (Ruffini's rule). We found that $1$ is a root of the polynomial

$1^6-2\cdot 1^4+1^2=0$

Now, divide the polynomial by the root we found $\left(x-1\right)$ 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 $1$. Add the result to the second coefficient and then multiply this by $1$ and so on

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

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 $\left(x-1\right)$

$\int\frac{x+8}{\left(x^{5}+x^{4}-x^{3}-x^{2}\right)\left(x-1\right)}dx$

We can factor the polynomial $\left(x^{5}+x^{4}-x^{3}-x^{2}\right)$ 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 $0$

$1$

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

$1$

The possible roots $\pm\frac{p}{q}$ of the polynomial $\left(x^{5}+x^{4}-x^{3}-x^{2}\right)$ will then be

$\pm5,\:\pm4,\:\pm3,\:\pm2,\:\pm1$

We can factor the polynomial $\left(x^{5}+x^{4}-x^{3}-x^{2}\right)$ using synthetic division (Ruffini's rule). We found that $1$ is a root of the polynomial

$1^{5}+1^{4}-1\cdot 1^{3}-1\cdot 1^{2}=0$

Now, divide the polynomial by the root we found $\left(x-1\right)$ 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 $1$. Add the result to the second coefficient and then multiply this by $1$ and so on

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

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 $\left(x-1\right)$

$\int\frac{x+8}{\left(x^{4}+2x^{3}+x^{2}\right)\left(x-1\right)\cdot \left(x-1\right)}dx$

We can factor the polynomial $\left(x^{4}+2x^{3}+x^{2}\right)$ 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 $0$

$1$

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

$1$

The possible roots $\pm\frac{p}{q}$ of the polynomial $\left(x^{4}+2x^{3}+x^{2}\right)$ will then be

$\pm5,\:\pm4,\:\pm3,\:\pm2,\:\pm1$

We can factor the polynomial $\left(x^{4}+2x^{3}+x^{2}\right)$ using synthetic division (Ruffini's rule). We found that $-1$ is a root of the polynomial

${\left(-1\right)}^{4}+2\cdot {\left(-1\right)}^{3}+{\left(-1\right)}^{2}=0$

Now, divide the polynomial by the root we found $\left(x+1\right)$ 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 $-1$. Add the result to the second coefficient and then multiply this by $-1$ and so on

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

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 $\left(x+1\right)$

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

Factor the polynomial $\left(x^{3}+x^{2}\right)$ by it's greatest common factor (GCF): $x^2$

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

Rewrite the expression $\frac{x+8}{x^6-2x^4+x^2}$ inside the integral in factored form

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

Rewrite the fraction $\frac{x+8}{\left(x-1\right)^2\left(x+1\right)^2x^2}$ in $6$ simpler fractions using partial fraction decomposition

$\frac{x+8}{\left(x-1\right)^2\left(x+1\right)^2x^2}=\frac{A}{\left(x-1\right)^2}+\frac{B}{\left(x+1\right)^2}+\frac{C}{x^2}+\frac{D}{x-1}+\frac{F}{x+1}+\frac{G}{x}$
3

Find the values for the unknown coefficients: $A, B, C, D, F, G$. The first step is to multiply both sides of the equation from the previous step by $\left(x-1\right)^2\left(x+1\right)^2x^2$

$x+8=\left(x-1\right)^2\left(x+1\right)^2x^2\left(\frac{A}{\left(x-1\right)^2}+\frac{B}{\left(x+1\right)^2}+\frac{C}{x^2}+\frac{D}{x-1}+\frac{F}{x+1}+\frac{G}{x}\right)$
4

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

$1\left(x+8\right)=1\left(x-1\right)^2\left(x+1\right)^2x^2\left(\frac{A}{\left(x-1\right)^2}+\frac{B}{\left(x+1\right)^2}+\frac{C}{x^2}+\frac{D}{x-1}+\frac{F}{x+1}+\frac{G}{x}\right)$
5

Multiplying polynomials

$1\left(x+8\right)=\frac{1\left(x-1\right)^2\left(x+1\right)^2x^2A}{\left(x-1\right)^2}+\frac{1\left(x-1\right)^2\left(x+1\right)^2x^2B}{\left(x+1\right)^2}+\frac{1\left(x-1\right)^2\left(x+1\right)^2x^2C}{x^2}+\frac{1\left(x-1\right)^2\left(x+1\right)^2x^2D}{x-1}+\frac{1\left(x-1\right)^2\left(x+1\right)^2x^2F}{x+1}+\frac{1\left(x-1\right)^2\left(x+1\right)^2x^2G}{x}$
6

Simplifying

$1\left(x+8\right)=\left(x+1\right)^2x^2A+\left(x-1\right)^2x^2B+\left(x-1\right)^2\left(x+1\right)^2C+\left(x-1\right)\left(x+1\right)^2x^2D+\left(x-1\right)^2\left(x+1\right)x^2F+\left(x-1\right)^2\left(x+1\right)^2xG$
7

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

$\begin{matrix}9=4A&\:\:\:\:\:\:\:(x=1) \\ 7=4B&\:\:\:\:\:\:\:(x=-1) \\ 10=36A+4B+9C+36D+12F+18G&\:\:\:\:\:\:\:(x=2) \\ 6=4A+36B+9C-12D-36F-18G&\:\:\:\:\:\:\:(x=-2) \\ 11=144A+36B+64C+288D+144F+192G&\:\:\:\:\:\:\:(x=3) \\ 5=36A+144B+64C-144D-288F-192G&\:\:\:\:\:\:\:(x=-3)\end{matrix}$
8

Proceed to solve the system of linear equations

$\begin{matrix}4A & + & 0B & + & 0C & + & 0D & + & 0F & + & 0G & =9 \\ 0A & + & 4B & + & 0C & + & 0D & + & 0F & + & 0G & =7 \\ 36A & + & 0B & + & 0C & + & 36D & + & 12F & + & 18G & =10 \\ 0A & + & 36B & + & 9C & - & 12D & - & 36F & - & 18G & =6 \\ 144A & + & 36B & + & 64C & + & 288D & + & 144F & + & 192G & =11 \\ 36A & + & 144B & + & 64C & - & 144D & - & 288F & - & 192G & =5\end{matrix}$
9

Rewrite as a coefficient matrix

$\left(\begin{matrix}4 & 0 & 0 & 0 & 0 & 0 & 9 \\ 0 & 4 & 0 & 0 & 0 & 0 & 7 \\ 36 & 0 & 0 & 36 & 12 & 18 & 10 \\ 0 & 36 & 9 & -12 & -36 & -18 & 6 \\ 144 & 36 & 64 & 288 & 144 & 192 & 11 \\ 36 & 144 & 64 & -144 & -288 & -192 & 5\end{matrix}\right)$
10

Reducing the original matrix to a identity matrix using Gaussian Elimination

$\left(\begin{matrix}1 & 0 & 0 & 0 & 0 & 0 & \frac{9}{4} \\ 0 & 1 & 0 & 0 & 0 & 0 & \frac{7}{4} \\ 0 & 0 & 1 & 0 & 0 & 0 & \frac{32}{37} \\ 0 & 0 & 0 & 1 & 0 & 0 & -2.943243 \\ 0 & 0 & 0 & 0 & 1 & 0 & 2.714414 \\ 0 & 0 & 0 & 0 & 0 & 1 & \frac{49}{370}\end{matrix}\right)$
11

The integral of $\frac{x+8}{\left(x-1\right)^2\left(x+1\right)^2x^2}$ in decomposed fraction equals

$\int\left(\frac{9}{4\left(x-1\right)^2}+\frac{7}{4\left(x+1\right)^2}+\frac{32}{37x^2}+\frac{-2.943243}{x-1}+\frac{2.714414}{x+1}+\frac{\frac{49}{370}}{x}\right)dx$
12

Expand the integral $\int\left(\frac{9}{4\left(x-1\right)^2}+\frac{7}{4\left(x+1\right)^2}+\frac{32}{37x^2}+\frac{-2.943243}{x-1}+\frac{2.714414}{x+1}+\frac{\frac{49}{370}}{x}\right)dx$ into $6$ integrals using the sum rule for integrals, to then solve each integral separately

$\int\frac{9}{4\left(x-1\right)^2}dx+\int\frac{7}{4\left(x+1\right)^2}dx+\int\frac{32}{37x^2}dx+\int\frac{-2.943243}{x-1}dx+\int\frac{2.714414}{x+1}dx+\int\frac{\frac{49}{370}}{x}dx$

Take the constant $\frac{1}{4}$ out of the integral

$\frac{1}{4}\int\frac{9}{\left(x-1\right)^2}dx$

Apply the formula: $\int\frac{n}{\left(x+a\right)^c}dx$$=\frac{-n}{\left(c-1\right)\left(x+a\right)^{\left(c-1\right)}}+C$, where $a=-1$, $c=2$ and $n=9$

$\frac{-9}{\left(2-1\right)\cdot 4\left(x-1\right)^{\left(2-1\right)}}$

Simplify the expression inside the integral

$\frac{-9}{4\left(x-1\right)}$
13

The integral $\int\frac{9}{4\left(x-1\right)^2}dx$ results in: $\frac{-9}{4\left(x-1\right)}$

$\frac{-9}{4\left(x-1\right)}$

Take the constant $\frac{1}{4}$ out of the integral

$\frac{1}{4}\int\frac{7}{\left(x+1\right)^2}dx$

Apply the formula: $\int\frac{n}{\left(x+a\right)^c}dx$$=\frac{-n}{\left(c-1\right)\left(x+a\right)^{\left(c-1\right)}}+C$, where $a=1$, $c=2$ and $n=7$

$\frac{-7}{\left(2-1\right)\cdot 4\left(x+1\right)^{\left(2-1\right)}}$

Simplify the expression inside the integral

$\frac{-7}{4\left(x+1\right)}$
14

The integral $\int\frac{7}{4\left(x+1\right)^2}dx$ results in: $\frac{-7}{4\left(x+1\right)}$

$\frac{-7}{4\left(x+1\right)}$

Take the constant $\frac{1}{37}$ out of the integral

$\frac{1}{37}\int\frac{32}{x^2}dx$

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

$\frac{1}{37}\int32x^{-2}dx$

The integral of a function times a constant ($32$) is equal to the constant times the integral of the function

$\frac{32}{37}\int x^{-2}dx$

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 $-2$

$-\frac{32}{37}x^{-1}$

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

$\frac{-32}{37x^{1}}$

Any expression to the power of $1$ is equal to that same expression

$\frac{-32}{37x}$
15

The integral $\int\frac{32}{37x^2}dx$ results in: $\frac{-32}{37x}$

$\frac{-32}{37x}$

The integral of a function times a constant ($-2.943243$) is equal to the constant times the integral of the function

$-2.943243\int\frac{1}{-1+x}dx$

Apply the formula: $\int\frac{n}{x+b}dx$$=nsign\left(x\right)\ln\left(x+b\right)+C$, where $b=-1$ and $n=1$

$-2.943243\ln\left(x-1\right)$
16

The integral $\int\frac{-2.943243}{x-1}dx$ results in: $-2.943243\ln\left(x-1\right)$

$-2.943243\ln\left(x-1\right)$

The integral of a function times a constant ($2.714414$) is equal to the constant times the integral of the function

$2.714414\int\frac{1}{1+x}dx$

Apply the formula: $\int\frac{n}{x+b}dx$$=nsign\left(x\right)\ln\left(x+b\right)+C$, where $b=1$ and $n=1$

$2.714414\ln\left(x+1\right)$
17

The integral $\int\frac{2.714414}{x+1}dx$ results in: $2.714414\ln\left(x+1\right)$

$2.714414\ln\left(x+1\right)$

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

$\frac{49}{370}\ln\left(x\right)$
18

The integral $\int\frac{\frac{49}{370}}{x}dx$ results in: $\frac{49}{370}\ln\left(x\right)$

$\frac{49}{370}\ln\left(x\right)$
19

Gather the results of all integrals

$\frac{-9}{4\left(x-1\right)}+\frac{-7}{4\left(x+1\right)}+\frac{-32}{37x}-2.943243\ln\left(x-1\right)+2.714414\ln\left(x+1\right)+\frac{49}{370}\ln\left(x\right)$
20

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

$\frac{-9}{4\left(x-1\right)}+\frac{-7}{4\left(x+1\right)}+\frac{-32}{37x}-2.943243\ln\left(x-1\right)+2.714414\ln\left(x+1\right)+\frac{49}{370}\ln\left(x\right)+C_0$

Final Answer

$\frac{-9}{4\left(x-1\right)}+\frac{-7}{4\left(x+1\right)}+\frac{-32}{37x}-2.943243\ln\left(x-1\right)+2.714414\ln\left(x+1\right)+\frac{49}{370}\ln\left(x\right)+C_0$

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Function Plot

Plotting: $\frac{-9}{4\left(x-1\right)}+\frac{-7}{4\left(x+1\right)}+\frac{-32}{37x}-2.943243\ln\left(x-1\right)+2.714414\ln\left(x+1\right)+\frac{49}{370}\ln\left(x\right)+C_0$

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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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