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### Difficult Problems

1

Solved example of gaussian elimination

$\int\frac{5x^2+14x+10}{\left(x+2\right)\left(x+1\right)^2}dx$
2

Rewrite the fraction $\frac{5x^2+14x+10}{\left(x+2\right)\left(x+1\right)^2}$ in $3$ simpler fractions using partial fraction decomposition

$\frac{5x^2+14x+10}{\left(x+2\right)\left(x+1\right)^2}=\frac{A}{x+2}+\frac{B}{\left(x+1\right)^2}+\frac{C}{x+1}$
3

Find the values of the unknown coefficients. The first step is to multiply both sides of the equation by $\left(x+2\right)\left(x+1\right)^2$

$5x^2+14x+10=\left(x+2\right)\left(x+1\right)^2\left(\frac{A}{x+2}+\frac{B}{\left(x+1\right)^2}+\frac{C}{x+1}\right)$
4

Multiplying polynomials

$5x^2+14x+10=\frac{A\left(x+2\right)\left(x+1\right)^2}{x+2}+\frac{B\left(x+2\right)\left(x+1\right)^2}{\left(x+1\right)^2}+\frac{C\left(x+2\right)\left(x+1\right)^2}{x+1}$
5

Simplifying

$5x^2+14x+10=A\left(x+1\right)^2+B\left(x+2\right)+C\left(x+2\right)\left(x+1\right)$
6

Expand the polynomial

$5x^2+14x+10=A\left(x+1\right)^2+Bx+2B+C\left(x\left(x+2\right)+x+2\right)$
7

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

$\begin{matrix}1=B&\:\:\:\:\:\:\:(x=-1) \\ 2=A&\:\:\:\:\:\:\:(x=-2) \\ 29=6C+3B+4A&\:\:\:\:\:\:\:(x=1)\end{matrix}$
8

Proceed to solve the system of linear equations

$\begin{matrix}0A & + & 1B & + & 0C & =1 \\ 1A & + & 0B & + & 0C & =2 \\ 4A & + & 3B & + & 6C & =29\end{matrix}$
9

Rewrite as a coefficient matrix

$\left(\begin{matrix}0 & 1 & 0 & 1 \\ 1 & 0 & 0 & 2 \\ 4 & 3 & 6 & 29\end{matrix}\right)$
10

Reducing the original matrix to a identity matrix using Gaussian Elimination

$\left(\begin{matrix}1 & 0 & 0 & 2 \\ 0 & 1 & 0 & 1 \\ 0 & 0 & 1 & 3\end{matrix}\right)$
11

The integral of $\frac{5x^2+14x+10}{\left(x+2\right)\left(x+1\right)^2}$ in decomposed fraction equals

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

The integral of the sum of two or more functions is equal to the sum of their integrals

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

The integral of the sum of two or more functions is equal to the sum of their integrals

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

Simplifying

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

The integral of a constant by a function is equal to the constant multiplied by the integral of the function

$2\int\frac{1}{2+x}dx$

Solve the integral $\int\frac{1}{2+x}dx$ applying u-substitution. Let $u$ and $du$ be

$\begin{matrix}u=2+x \\ du=dx\end{matrix}$

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

$2\int\frac{1}{u}du$

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

$2\ln\left|u\right|$

Substitute $u$ back for it's value, $2+x$

$2\ln\left|2+x\right|$
13

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

$2\ln\left|2+x\right|$

Solve the integral $\int\frac{1}{\left(x+1\right)^2}dx$ applying u-substitution. Let $u$ and $du$ be

$\begin{matrix}u=x+1 \\ du=dx\end{matrix}$

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

$\int\frac{1}{u^2}du$

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

$\int u^{-2}du$

Apply the power rule for integration, $\displaystyle\int x^n dx=\frac{x^{n+1}}{n+1}$, where $n$ represents a constant function

$-u^{-1}$

Substitute $u$ back for it's value, $x+1$

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

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

$-\frac{1}{x+1}$

Multiplying the fraction and term

$\frac{-1}{x+1}$
14

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

$\frac{-1}{x+1}$

The integral of a constant by a function is equal to the constant multiplied by the integral of the function

$3\int\frac{1}{1+x}dx$

Solve the integral $\int\frac{1}{1+x}dx$ applying u-substitution. Let $u$ and $du$ be

$\begin{matrix}u=1+x \\ du=dx\end{matrix}$

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

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

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

$3\ln\left|u\right|$

Substitute $u$ back for it's value, $1+x$

$3\ln\left|1+x\right|$
15

The integral $\int\frac{3}{x+1}dx$ results in: $3\ln\left|1+x\right|$

$3\ln\left|1+x\right|$
16

After gathering the results of all integrals, the final answer is

$2\ln\left|2+x\right|+\frac{-1}{x+1}+3\ln\left|1+x\right|$
17

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

$2\ln\left|2+x\right|+\frac{-1}{x+1}+3\ln\left|1+x\right|+C_0$

$2\ln\left|2+x\right|+\frac{-1}{x+1}+3\ln\left|1+x\right|+C_0$