Calcolatrice di Matrices

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

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

Multiplying polynomials

Simplifying

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

Proceed to solve the system of linear equations

Rewrite as a coefficient matrix

Reducing the original matrix to a identity matrix using Gaussian Elimination

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

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

Take out the negative sign of all the terms of the numerator of the integral

Add and subtract $\displaystyle\left(\frac{b}{2a}\right)^2$

Factor the perfect square trinomial $x^2+x+\frac{1}{4}$

Calculate the power $\sqrt{\frac{1}{4}}$

Simplify the addition $\left(x+\frac{1}{2}\right)^2+1-\frac{1}{4}$

Multiply $1$ times $4$

Subtract the values $4$ and $-1$

We can solve the integral $\int\frac{x+1}{\left(x+\frac{1}{2}\right)^2+\frac{3}{4}}dx$ 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 $x+\frac{1}{2}$ 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 $\frac{u+\frac{1}{2}}{u^2+\frac{3}{4}}$ into $2$ simpler fractions with common denominator $u^2+\frac{3}{4}$

Expand the integral $\int\left(\frac{u}{u^2+\frac{3}{4}}+\frac{\frac{1}{2}}{u^2+\frac{3}{4}}\right)du$ into $2$ integrals using the sum rule for integrals, to then solve each integral separately

The integral of a function times a constant ($\frac{1}{2}$) is equal to the constant times the integral of the function

Multiply the fraction and term in $- \left(\frac{1}{2}\right)\int\frac{1}{u^2+\frac{3}{4}}du$

Factor the integral's denominator by $\frac{3}{4}$

Simplify the expression

We can solve the integral $-\int\frac{u}{u^2+\frac{3}{4}}du$ by applying integration method of trigonometric substitution using the substitution

Now, in order to rewrite $d\theta$ 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

Substituting in the original integral, we get

Simplifying

Simplify the expression

The integral of a function times a constant ($3$) is equal to the constant times the integral of the function

Multiply the fraction and term in $3\left(-\frac{2}{3}\right)\int\frac{1}{3+4u^2}du$

The integral of the tangent function is given by the following formula, $\displaystyle\int\tan(x)dx=-\ln(\cos(x))$

Any expression multiplied by $1$ is equal to itself

Express the variable $\theta$ in terms of the original variable $x$

Divide fractions $\frac{\frac{\sqrt{3}}{2}}{\sqrt{u^2+\frac{3}{4}}}$ with Keep, Change, Flip: $\frac{a}{b}\div c=\frac{a}{b}\div\frac{c}{1}=\frac{a}{b}\times\frac{1}{c}=\frac{a}{b\cdot c}$

Replace $u$ with the value that we assigned to it in the beginning: $x+\frac{1}{2}$

Solve the integral applying the substitution $v^2=\frac{4u^2}{3}$. Then, take the square root of both sides, simplifying we have

Now, in order to rewrite $du$ in terms of $dv$, we need to find the derivative of $v$. We need to calculate $dv$, we can do that by deriving the equation above

Isolate $du$ in the previous equation

After replacing everything and simplifying, the integral results in

Solve the integral by applying the formula $\displaystyle\int\frac{x'}{x^2+a^2}dx=\frac{1}{a}\arctan\left(\frac{x}{a}\right)$

Multiplying the fraction by $\arctan\left(v\right)$

Replace $v$ with the value that we assigned to it in the beginning: $\frac{2u}{\sqrt{3}}$

Replace $u$ with the value that we assigned to it in the beginning: $x+\frac{1}{2}$

Simplify the product by distributing $2$ to both terms