`int x / sqrt(9 + 8x^2 - x^4) dx` Find or evaluate the integral by completing the square

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marizi eNotes educator| Certified Educator

Recall  that` int f(x) dx = F(x) +C` where:

f(x) as the integrand function

F(x) as the antiderivative of f(x)

C as the constant of integration..

 For the given problem, the integral: `int x/sqrt(9+8x^2-x^4)dx`

does not yet resemble any formula from table of integrals.

 

To evaluate this, we are to apply u-substitution by letting:

`u = x^2` then `u^2 = x^4`  and  `du = 2x dx `  or `(du)/2 = x dx` .

Then the integral becomes:

`int x/sqrt(9+8x^2-x^4)dx =int x dx/sqrt(9+8x^2-x^4)`

                                      `=int ((du)/2)/sqrt(9+8u-u^4)`

Apply the  basic  property of integration: `int c f(x) dx = c int f(x) dx` to factor out ` 1/2` .

`int ((du)/2)/sqrt(9+8u-u^4) = 1/2int (du)/sqrt(9+8u-u^4)`

 The integral does not yet resembles any integration formula.

For further step, we apply  completing the square on the part: `9+8u-u^2` .

Completing the square:

Factoring out -1 from `9+8u-u^2` becomes: `(-1)(-9-8u^2 +u^2)` or `-(u^2 -8u-9)` .

`u^2 -8u-9` resembles `ax^2 +bx+c ` where:

`a=1` ,` b= -8` and `c=9` .

To complete the square we add and subtract `(-b/(2a))^2` .

Plug-in the value of `a=1` and `b=-8` in  `(-b/(2a))^2` :

`(-b/(2a))^2 =(-(-8)/(2*1))^2`

             `=(8/2)^2`

             ` =4^2`

             ` =16.`

Adding and subtracting -16 inside the ():

`-(u^2 -8u-9) =-(u^2 -8u-9+16-16)`

 To move out "-9" and "-16" outside the (), we distribute the negative sign or (-1).

` -(u^2 -8u-9+16-16) =-(u^2 -8u-9+16) +(-1)(-9)+ (-1)(-16) `  

                                         `=-(u^2 -8u-9+16) +9+ 16`

                                         `=-(u^2 -8u-9+16) +25`

Factor out the perfect square trinomial: `u^2 -8u+16 = (u-4)^2`

`-(u^2 -8u+16) + 16 = -(u-4)^2+25`

Then it shows that `9+8u-u^4 =-(u-4)^2+25`

                                               `=25-(u-4)^2 `

                                                ` = 5^2 -(u-4)^2`

Then,

`1/2 int (du)/sqrt(9+8u-u^4)= 1/2int (du)/sqrt(5^2-(u-4)^2)`

 The integral part resembles the basic integration formula for inverse sine function:

`int (du)/sqrt(a^2-u^2)= arcsin(u/a)+C`

Applying the formula, we get:

`1/2int (du)/sqrt(5^2-(u-4)^2) =1/2 arcsin ((u-4)/5) +C`

Plug-in `u =x^2`  for the final answer:

`int x/sqrt(9+8x^2-x^4)dx =1/2 arcsin ((x^2-4)/5) +C`