Calculus of a Single Variable, Chapter 5, 5.7, Section 5.7, Problem 42

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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