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`
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