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`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`

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