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# Gaussian Elimination Calculator

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###  Difficult Problems

1

Here, we show you a step-by-step solved example of gaussian elimination. This solution was automatically generated by our smart calculator:

$\int_{-\infty}^{\infty}\left(\frac{1}{2x^2\left(x-1\right)}\right)dx$
2

Take the constant $\frac{1}{2}$ out of the integral

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

Rewrite the fraction $\frac{1}{x^2\left(x-1\right)}$ in $3$ simpler fractions using partial fraction decomposition

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

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 $x^2\left(x-1\right)$

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

Multiplying polynomials

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

Simplifying

$1=\left(x-1\right)A+x^2B+x\left(x-1\right)C$

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

$\begin{matrix}1=-A&\:\:\:\:\:\:\:(x=0) \\ 1=B&\:\:\:\:\:\:\:(x=1) \\ 1=-2A+B+2C&\:\:\:\:\:\:\:(x=-1)\end{matrix}$

Proceed to solve the system of linear equations

$\begin{matrix} -1A & + & 0B & + & 0C & =1 \\ 0A & + & 1B & + & 0C & =1 \\ -2A & + & 1B & + & 2C & =1\end{matrix}$

Rewrite as a coefficient matrix

$\left(\begin{matrix}-1 & 0 & 0 & 1 \\ 0 & 1 & 0 & 1 \\ -2 & 1 & 2 & 1\end{matrix}\right)$

Reducing the original matrix to a identity matrix using Gaussian Elimination

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

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

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

Rewrite the fraction $\frac{1}{x^2\left(x-1\right)}$ in $3$ simpler fractions using partial fraction decomposition

$\frac{-1}{x^2}+\frac{1}{x-1}+\frac{-1}{x}$
4

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

$\frac{1}{2}\int\frac{-1}{x^2}dx+\frac{1}{2}\int\frac{1}{x-1}dx+\frac{1}{2}\int\frac{-1}{x}dx$
5

We can solve the integral $\int\frac{1}{x-1}dx$ 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 $x-1$ it's a good candidate for substitution. Let's define a variable $u$ and assign it to the choosen part

$u=x-1$

Differentiate both sides of the equation $u=x-1$

$du=\frac{d}{dx}\left(x-1\right)$

Find the derivative

$\frac{d}{dx}\left(x-1\right)$

The derivative of a sum of two or more functions is the sum of the derivatives of each function

$1$
6

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

$du=dx$
7

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

$\frac{1}{2}\int\frac{-1}{x^2}dx+\frac{1}{2}\int\frac{1}{u}du+\frac{1}{2}\int\frac{-1}{x}dx$

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

$\frac{1}{2}\int-x^{-2}dx$

The integral of a function times a constant ($-1$) is equal to the constant times the integral of the function

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

Multiply the fraction and term in $-\left(\frac{1}{2}\right)\int x^{-2}dx$

$-\frac{1}{2}\int x^{-2}dx$

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 $-2$

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

Simplify the expression

$\frac{1}{2x}$
8

The integral $\frac{1}{2}\int\frac{-1}{x^2}dx$ results in: $\frac{1}{2x}$

$\frac{1}{2x}$

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

$\frac{1}{2}\ln\left|u\right|$

Replace $u$ with the value that we assigned to it in the beginning: $x-1$

$\frac{1}{2}\ln\left|x-1\right|$
9

The integral $\frac{1}{2}\int\frac{1}{u}du$ results in: $\frac{1}{2}\ln\left|x-1\right|$

$\frac{1}{2}\ln\left|x-1\right|$

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

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

Multiply the fraction and term in $-\left(\frac{1}{2}\right)\ln\left|x\right|$

$-\frac{1}{2}\ln\left|x\right|$
10

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

$-\frac{1}{2}\ln\left|x\right|$
11

Gather the results of all integrals

$\frac{1}{2x}+\frac{1}{2}\ln\left|x-1\right|-\frac{1}{2}\ln\left|x\right|$
12

Add the initial limits of integration

$\left[\left(\frac{1}{2x}+\frac{1}{2}\ln\left(x-1\right)-\frac{1}{2}\ln\left(x\right)\right)\right]_{- \infty }^{\infty }$
13

Replace the integral's limit by a finite value

$\lim_{c\to{- \infty }}\left(\left[\left(\frac{1}{2x}+\frac{1}{2}\ln\left|x-1\right|-\frac{1}{2}\ln\left|x\right|\right)\right]_{c}^{\infty }\right)$
14

Evaluate the definite integral

$\lim_{c\to{- \infty }}\left(\frac{1}{2\cdot \infty }+\frac{1}{2}\ln\left|\infty -1\right|-\frac{1}{2}\ln\left|\infty \right|-\left(\frac{1}{2c}+\frac{1}{2}\ln\left|c-1\right|-\frac{1}{2}\ln\left|c\right|\right)\right)$

The natural log of infinity is equal to infinity, $\lim_{x\to\infty}\ln(x)=\infty$

$\lim_{c\to{- \infty }}\left(\frac{1}{2\cdot \infty }+\frac{1}{2}\ln\left(\infty -1\right)-\frac{1}{2}\cdot \infty -\left(\frac{1}{2c}+\frac{1}{2}\ln\left(c-1\right)-\frac{1}{2}\ln\left(c\right)\right)\right)$

Any expression multiplied by infinity tends to infinity, in other words: $\infty\cdot(\pm n)=\pm\infty$, if $n\neq0$

$\lim_{c\to{- \infty }}\left(\frac{1}{\infty }+\frac{1}{2}\ln\left(\infty -1\right)-\frac{1}{2}\cdot \infty -\left(\frac{1}{2c}+\frac{1}{2}\ln\left(c-1\right)-\frac{1}{2}\ln\left(c\right)\right)\right)$

Any expression multiplied by infinity tends to infinity, in other words: $\infty\cdot(\pm n)=\pm\infty$, if $n\neq0$

$\lim_{c\to{- \infty }}\left(\frac{1}{\infty }+\frac{1}{2}\ln\left(\infty -1\right)- \infty -\left(\frac{1}{2c}+\frac{1}{2}\ln\left(c-1\right)-\frac{1}{2}\ln\left(c\right)\right)\right)$

Infinity plus any algebraic expression is equal to infinity

$\lim_{c\to{- \infty }}\left(\frac{1}{\infty }+\frac{1}{2}\ln\left(\infty \right)- \infty -\left(\frac{1}{2c}+\frac{1}{2}\ln\left(c-1\right)-\frac{1}{2}\ln\left(c\right)\right)\right)$

The natural log of infinity is equal to infinity, $\lim_{x\to\infty}\ln(x)=\infty$

$\lim_{c\to{- \infty }}\left(\frac{1}{\infty }+\frac{1}{2}\cdot \infty - \infty -\left(\frac{1}{2c}+\frac{1}{2}\ln\left(c-1\right)-\frac{1}{2}\ln\left(c\right)\right)\right)$

Any expression multiplied by infinity tends to infinity, in other words: $\infty\cdot(\pm n)=\pm\infty$, if $n\neq0$

$\lim_{c\to{- \infty }}\left(\frac{1}{\infty }+\infty - \infty -\left(\frac{1}{2c}+\frac{1}{2}\ln\left(c-1\right)-\frac{1}{2}\ln\left(c\right)\right)\right)$

Any expression divided by infinity is equal to zero

$\lim_{c\to{- \infty }}\left(\infty - \infty -\left(\frac{1}{2c}+\frac{1}{2}\ln\left(c-1\right)-\frac{1}{2}\ln\left(c\right)\right)\right)$

Infinity minus infinity is an indeterminate form

indeterminate
15

Evaluate the resulting limits of the integral

indeterminate
16

When the limits of the integral do not exist, it is said that the integral is divergent

The integral diverges.

##  Final answer to the problem

The integral diverges.

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