Student Question

# int sqrt(x)/(x-4) dx Use substitution and partial fractions to find the indefinite integral

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