Integral of (x-2)/(x(5x-14+x^2))

\int\frac{x-2}{x\cdot\left(x^2+5x-14\right)}dx

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Answer

$-\frac{1}{7}\ln\left|7+x\right|+\frac{1}{7}\ln\left|x\right|+C_0$

Step by step solution

Problem

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

Factor the trinomial $\left(-14+5x+x^2\right)$ finding two numbers that multiply to form $-14$ and added form $5$

$\begin{matrix}\left(-2\right)\left(7\right)=-14\\ \left(-2\right)+\left(7\right)=5\end{matrix}$
2

Thus

$\int\frac{x-2}{x\left(7+x\right)\left(x-2\right)}dx$
3

Simplifying the fraction by $x-2$

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

Using partial fraction decomposition, the fraction $\frac{1}{x\left(7+x\right)}$ can be rewritten as

$\frac{1}{x\left(7+x\right)}=\frac{A}{7+x}+\frac{B}{x}$
5

Now we need to find the values of the unknown coefficients. The first step is to multiply both sides of the equation by $x\left(7+x\right)$

$1=x\left(\frac{A}{7+x}+\frac{B}{x}\right)\left(7+x\right)$
6

Multiplying polynomials

$1=\frac{Ax\left(7+x\right)}{7+x}+\frac{Bx\left(7+x\right)}{x}$
7

Simplifying

$1=Ax+B\left(7+x\right)$
8

Expand the polynomial

$1=Ax+7B+Bx$
9

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

$\begin{matrix}1=7B&\:\:\:\:\:\:\:(x=0) \\ 1=-7A+7B-7B&\:\:\:\:\:\:\:(x=-7)\end{matrix}$
10

Proceed to solve the system of linear equations

$\begin{matrix}0A & + & 7B & =1 \\ -7A & + & 0B & =1\end{matrix}$
11

Rewrite as a coefficient matrix

$\left(\begin{matrix}0 & 7 & 1 \\ -7 & 0 & 1\end{matrix}\right)$
12

Reducing the original matrix to a identity matrix using Gaussian Elimination

$\left(\begin{matrix}1 & 0 & -\frac{1}{7} \\ 0 & 1 & \frac{1}{7}\end{matrix}\right)$
13

The decomposed integral equivalent is

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

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

$\int\frac{-\frac{1}{7}}{7+x}dx+\int\frac{\frac{1}{7}}{x}dx$
15

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

$\int\frac{-\frac{1}{7}}{7+x}dx+\frac{1}{7}\ln\left|x\right|$
16

Apply the formula: $\int\frac{n}{b+x}dx$$=n\ln\left|b+x\right|$, where $b=7$ and $n=-\frac{1}{7}$

$\frac{1}{7}\ln\left|x\right|-\frac{1}{7}\ln\left|7+x\right|$
17

Add the constant of integration

$-\frac{1}{7}\ln\left|7+x\right|+\frac{1}{7}\ln\left|x\right|+C_0$

Answer

$-\frac{1}{7}\ln\left|7+x\right|+\frac{1}{7}\ln\left|x\right|+C_0$

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Problem Analysis

Main topic:

Integrals by partial fraction expansion

Time to solve it:

0.41 seconds

Views:

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