Calculus of a Single Variable, Chapter 8, 8.5, Section 8.5, Problem 19

int x^2/(x^4-2x^2-8)dx
To solve using partial fraction method, the denominator of the integrand should be factored.
x^2/(x^4-2x^2-8)=x^2/((x-2)(x+2)(x^2+2))
If the factor in the denominator is linear, its partial fraction has a form A/(ax+b) .
If the factor is quadratic, its partial fraction is in the form (Ax+B)/(ax^2+bx+c) .
So, expressing the integrand as sum of fractions, it becomes:
x^2/((x-2)(x+2)(x^2+2)) = A/(x-2) + B/(x + 2) + (Cx + D)/(x^2+2)
To determine the values of A, B, C and D, multiply both sides by the LCD of the fractions present
(x-2)(x+2)(x^2+2)*x^2/((x-2)(x+2)(x^2+2)) = (A/(x-2) + B/(x + 2) + (Cx + D)/(x^2+2))*(x-2)(x+2)(x^2+2)
x^2=A(x+2)(x^2+2) + B(x-2)(x^2+2) + (Cx+D)(x-2)(x+2)
Then, assign values to x in which either x-2,x+2 orx^2+2will become zero.
So plug-in x=2 to get the value of A.
2^2=A(2+2)(2^2+2) + B(2-2)(2^2+2) + (C*2+D)(2-2)(2+2)
4=A(24)+B(0)+(2C+D)(0)
4=24A
1/6=A
Plug-inx=-2 to get the value of B.
(-2)^2=A(-2+2)((-2)^2+2) + B(-2-2)((-2)^2+2) + (C(-2)+D)(-2-2)(-2+2)
4=A(0)+B(-24) + (-2C+D)(0)
4=-24B
-1/6=B
To solve for D, plug-in the values of A and B. Also, set the value of x to zero.
0^2 =1/6(0+2)(0^2+2) + (-1/6)(0-2)(0^2+2) + (C(0)+D)(0-2)(0+2)
0=2/3+2/3-4D
0=4/3-4D
4D=4/3
D=1/3
To solve for C, plug-in the values of A, B and D. Also, assign any value to x. Let it be x=1.
1^2=1/6(1+2)(1^2+2)+( -1/6)(1-2)(1^2+2) + (C(1)+1/3)(1-2)(1+2)
1=3/2 +1/2+(C+1/3)(-3)
1=3/2+1/2-3C -1
1=1-3C
0=-3C
0=C
So the partial fraction decomposition of the integrand is:
int x^2/(x^4-2x^2-8)dx
=intx^2/((x-2)(x+2)(x^2+2))dx
=int (1/(6(x-2)) - 1/(6(x + 2)) + 1/(3(x^2+2)))dx
Expressing it as three integrals, it becomes
= int 1/(6(x-2))dx - int 1/(6(x+2))dx + int 1/(3(x^2+2))dx
= 1/6 int1/(x-2)dx - 1/6 int1/(x+2)dx + 1/3 int 1/(x^2+2)dx
For the first and second integral, apply the formula int 1/u du = ln|u|+C .
And for the third integral, the formula isint 1/(u^2+a^2)du =1/a tan^(-1)(u/a)+C .
=1/6ln|x-2| - 1/6ln|x+2| + 1/3*1/sqrt2 tan^(-1)(x/sqrt2)+C
=1/6ln|x-2| - 1/6ln|x+2| + 1/(3sqrt2) tan^(-1)(x/sqrt2)+C
=1/6ln|x-2| - 1/6ln|x+2| + sqrt2/6 tan^(-1)((xsqrt2)/2)+C

Therefore, int x^2/(x^4-2x^2-8)dx=1/6ln|x-2| - 1/6ln|x+2| + sqrt2/6 tan^(-1)((xsqrt2)/2)+C .