How to solve this integral ?
int x^2/(x^4 + x^2 -2) dx
2 Answers
Explanation:
In the denominator, letting
y^2 + y -2
(y+ 2)(y - 1)
Thus,
(x^2 + 2)(x^2 - 1)
(x^2+ 2)(x + 1)(x - 1)
Now we can integrate using partial fractions:
A/(x + 1) + B/(x - 1) + C/(x^2 + 2) = x^2/((x^2 + 2)(x + 1)(x - 1))
A(x - 1)(x^2 + 2) + B(x + 1)(x^2 + 2) + C(x + 1)(x - 1) = x^2
A(x^3 - x^2 + 2x - 2) + B(x^3 + x^2 + 2x + 2) + C(x^2 - 1) = x^2
Ax^3 - Ax^2 + 2Ax - 2A + Bx^3 + Bx^2 + 2Bx + 2B + Cx^2 - C = x^2
Therefore,
{(A + B = 0), (B - A + C = 1), (2A + 2B = 0), (-2A+ 2B - C = 0):}
The first and third equations are the same. Therefore our new system becomes
{(A + B = 0), (B - A + C = 1), (2B - 2A - C = 0):}
We can add equation two to equation three eliminating variable
{(A+ B = 0), (3B - 3A = 1):}
You can easily solve this by elimination.
{(3A + 3B = 0), (-3A + 3B = 1):}
Therefore,
Hence the integral becomes
I = int -1/(6(x + 1)) + 1/(6(x - 1)) + 2/(3(x^2 + 2))
The first and second terms are easily integrated. The third can be rearranged to the form
I = 1/6ln|x - 1| - 1/6ln|x +1| + sqrt(2)arctan(x/sqrt(2))/3 + C
If you differentiate this you will get the original integral.
Hopefully this helps!
Explanation:
Letting
by Heaviside's Method,
Replacing
HSBC244 has already derived!.