Calculus of a Single Variable, Chapter 8, 8.5, Section 8.5, Problem 33

Indefinite integral are written in the form of int f(x) dx = F(x) +C
where: f(x) as the integrand
F(x) as the anti-derivative function of f(x)
C as the arbitrary constant known as constant of integration
To evaluate the integral problem: int sqrt(x)/(x-4)dx , we may apply u-substitution by letting:
u=sqrt(x) then u^2 =x and 2u du = dx
Plug-in the values, we get:
int sqrt(x)/(x-4)dx=int u/(u^2-4)* 2udu
= int (2u^2)/(u^2-4)du
To simplify, we may apply long division:(2u^2)/(u^2-4) =2 +8/(u^2-4)
To expand 8/(u^2-4) , we may apply partial fraction decomposition.
The pattern on setting up partial fractions will depend on the factors of the denominator. The factored form for the difference of perfect squares: (u^2-4)= (u-2)(u+2) .

For the linear factor (u-2) , we will have partial fraction: A/(u-2) .
For the linear factor (u+2) , we will have partial fraction: B/(u+2) .
The rational expression becomes:
8/(u^2-4) =A/(u-2) +B/(u+2)
Multiply both side by the LCD =(u-2)(u+2) .
(8/(u^2-4)) *(u-2)(u+2)=(A/(u-2) +B/(u+2)) *(u-2)(u+2)
8=A(u+2) +B(u-2)
We apply zero-factor property on (u-2)(u+2) to solve for values we can assign on u.
u-2=0 then u=2
u+2 =0 then u =-2
To solve for A , we plug-in u=2 :
8=A(2+2) +B(2-2)
8 =4A+0
8=4A
8/(4) = (4A)/4
A = 2
To solve for B , we plug-in u=-2 :
8=A(-2+2) +B(-2-2)
8 =0 -4B
8=-4B
8/(-4) = (-4B)/(-4)
B = -2
Plug-in A = 2 and B =-2 , we get the partial fraction decomposition:
8/(u^2-4)=2/(u-2) -2/(u+2)
Then the integral becomes:
int (2u^2)/(u^2-4)du= int [2+8/(u^2-4)]du
=int [2 +2/(u-2) -2/(u+2)]du
Apply the basic integration property: int (u+-v+-w) dx = int (u) dx +- int (v) dx+- int (w) dx .
int [2 +2/(u-2) -2/(u+2)]du =int 2du +int 2/(u-2) du int -2/(u+2)du
For the first integral, we may apply basic integration property: int a dx = ax+C.
int 2 du = 2u
For the second and third integral, we may apply integration formula for logarithm: int 1/u du = ln|u|+C .
int 2/(u-2) du =2ln|u-2|
int 2/(u+2) du =2ln|u+2|
Combining the results, we get:
int (2u^2)/(u^2-4)du = 2u +2ln|u-2| -2ln|u+2| +C
Apply logarithm property: n*ln|x| = ln|x^n| and ln|x| - ln|y| = ln|x/y|
int (2u^2)/(u^2-4)du = 2u + ln|(u-2)^2| - ln|(u+2)^2| +C
= 2u + ln|(u-2)^2/(u+2)^2| +C
Plug-in u =sqrt(x) on 2u + ln|(u-2)^2/(u+2)^2| +C , we get the indefinite integral as:
int sqrt(x)/(x-4)dx =2sqrt(x) +ln|(sqrt(x)-2)^2/(sqrt(x)+2)^2| +C
OR 2sqrt(x) +ln|(x-4sqrt(x)+4)/(x+4sqrt(x)+4)| +C