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!.