# What is the improper integrals 1/(x^2-16) dx from 0 to 4 ? .

Jun 1, 2018

The integral diverges.
color(blue)[int_0^4(1)/(x^2-16)*dx=lim_(brarr4)int_0^b(1)/(x^2-16)*dx=-1/4*lim_(brarr4)[ln|csctheta+cottheta|]_0^b

#### Explanation:

show below:

${\int}_{0}^{4} \frac{1}{{x}^{2} - 16} \cdot \mathrm{dx} = {\lim}_{b \rightarrow 4} {\int}_{0}^{b} \frac{1}{{x}^{2} - 16} \cdot \mathrm{dx}$

The integral in the top can be integrated using two ways but I will use one of them: trigonometric substitution.

Suppose:

$x = 4 \cdot \sec \theta$

$\mathrm{dx} = 4 \cdot \sec \theta \cdot \tan \theta \cdot d \theta$

$\left({x}^{2} - 16\right) = 16 {\sec}^{2} \theta - 16 = 16 \cdot \left({\sec}^{2} \theta - 1\right) = 16 \cdot {\tan}^{2} \theta$

${\lim}_{b \rightarrow 4} {\int}_{0}^{b} \frac{4 \cdot \sec \theta \cdot \tan \theta \cdot d \left(\theta\right)}{16 \cdot {\tan}^{2} \theta} =$

$\frac{1}{4} \cdot {\lim}_{b \rightarrow 4} {\int}_{0}^{b} \frac{\sec \theta \cdot d \left(\theta\right)}{\tan \theta} =$

$\frac{1}{4} \cdot {\lim}_{b \rightarrow 4} {\int}_{0}^{b} \left(\frac{1}{\sin \theta}\right) = \frac{1}{4} \cdot {\lim}_{b \rightarrow 4} {\int}_{0}^{b} \csc \theta \cdot d \left(\theta\right)$

$\frac{1}{4} \cdot {\lim}_{b \rightarrow 4} {\int}_{0}^{b} \csc \theta \cdot \frac{\left(\csc \theta + \cot \theta\right)}{\left(\csc \theta + \cot \theta\right)} \cdot d \left(\theta\right)$

$\frac{1}{4} \cdot {\lim}_{b \rightarrow 4} {\int}_{0}^{b} \frac{\left({\csc}^{2} \theta + \csc \theta \cdot \cot \theta\right)}{\left(\csc \theta + \cot \theta\right)} \cdot d \left(\theta\right)$

$- \frac{1}{4} \cdot {\lim}_{b \rightarrow 4} {\left[\ln | \csc \theta + \cot \theta |\right]}_{0}^{b}$

Jun 1, 2018

The integral diverges.

#### Explanation:

Perform the decomposition into partial fractions

$\frac{1}{{x}^{2} - 16} = \frac{1}{\left(x + 4\right) \left(x - 4\right)} = \frac{A}{x + 4} + \frac{B}{x - 4}$

$= \frac{A \left(x - 4\right) + B \left(x + 4\right)}{\left(x + 4\right) \left(x - 4\right)}$

The denominators are the same, compare the numerators

$1 = A \left(x - 4\right) + B \left(x + 4\right)$

Let $x = - 4$, $\implies$, $1 = - 8 A$, $\implies$, $A = - \frac{1}{8}$

Let $x = 4$, $\implies$, $1 = 8 B$, $\implies$, $B = \frac{1}{8}$

Therefore,

$\frac{1}{{x}^{2} - 16} = \frac{- \frac{1}{8}}{x + 4} + \frac{\frac{1}{8}}{x - 4}$

The indefinite integral is

$\int \frac{1 \mathrm{dx}}{{x}^{2} - 16} = \int \frac{- \frac{1}{8} \mathrm{dx}}{x + 4} + \int \frac{\frac{1}{8} \mathrm{dx}}{x - 4}$

$= - \frac{1}{8} \ln \left(| x + 4 |\right) + \frac{1}{8} \ln \left(| x - 4 |\right) + C$

Compute the boundaries :

${\lim}_{x \to {0}^{+}} - \frac{1}{8} \ln \left(| x + 4 |\right) + \frac{1}{8} \ln \left(| x - 4 |\right) = 0$

${\lim}_{x \to {4}^{-}} - \frac{1}{8} \ln \left(| x + 4 |\right) + \frac{1}{8} \ln \left(| x - 4 |\right) = - \infty$

Therefore,

${\int}_{0}^{4} \frac{\mathrm{dx}}{{x}^{2} - 16} = \text{ diverges }$