Question

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

Answer 1

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(1)/(x^2-16)*dx=lim_(brarr4)int_0^b(1)/(x^2-16)*dx`

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

Suppose:

`x=4*sectheta`

`dx=4*sectheta * tantheta* d theta`

`(x^2-16)= 16sec^2 theta - 16=16*(sec^2theta-1)=16*tan^2theta`

`lim_(brarr4)int_0^b(4*sectheta*tantheta*d(theta))/(16*tan^2theta)=`

`1/4*lim_(brarr4)int_0^b(sectheta*d(theta))/(tantheta)=`

`1/4*lim_(brarr4)int_0^b(1/(sintheta))=1/4*lim_(brarr4)int_0^bcsctheta*d(theta)`

`1/4*lim_(brarr4)int_0^bcsctheta*[(csctheta+cottheta)]/[(csctheta+cottheta)]*d(theta)`

`1/4*lim_(brarr4)int_0^b[(csc^2theta+csctheta*cottheta)]/[(csctheta+cottheta)]*d(theta)`

`-1/4*lim_(brarr4)[ln|csctheta+cottheta|]_0^b`

Answer 2

The integral diverges.

Explanation:

Perform the decomposition into partial fractions

`1/(x^2-16)=1/((x+4)(x-4))=A/(x+4)+B/(x-4)`

`=(A(x-4)+B(x+4))/((x+4)(x-4))`

The denominators are the same, compare the numerators

`1=A(x-4)+B(x+4)`

Let `x=-4`, `=>`, `1=-8A`, `=>`, `A=-1/8`

Let `x=4`, `=>`, `1=8B`, `=>`, `B=1/8`

Therefore,

`1/(x^2-16)=(-1/8)/(x+4)+(1/8)/(x-4)`

The indefinite integral is

`int(1dx)/(x^2-16)=int(-1/8dx)/(x+4)+int(1/8dx)/(x-4)`

`=-1/8ln(|x+4|)+1/8ln(|x-4|)+C`

Compute the boundaries :

`lim_(x->0^+)-1/8ln(|x+4|)+1/8ln(|x-4|)=0`

`lim_(x->4^-)-1/8ln(|x+4|)+1/8ln(|x-4|)=-oo`

Therefore,

`int_0^4(dx)/(x^2-16)= " diverges "`