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

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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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