Question

How to solve this integral ?

`int x^2/(x^4 + x^2 -2) dx`

Answer 1

`I = 1/6ln|x - 1| - 1/6ln|x + 1| + sqrt(2)arctan(x/sqrt(2))/3 + C`

Explanation:

In the denominator, letting `y=x^2`, we can rewrite as

`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 `C`.

`{(A+ B = 0), (3B - 3A = 1):}`

You can easily solve this by elimination.

`{(3A + 3B = 0), (-3A + 3B = 1):}`

Therefore, `6B = 1` and `B = 1/6`. It follows that `A = -1/6`, and that `C = 1- 1/3 = 2/3`

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 `1/(x^2 + 1)` and hence integrated using the arctangent identity.

`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!

Answer 2

`sqrt2/3arc tan(x/sqrt2)+1/6ln|(x-1)/(x+1)|+C`.

Explanation:

Letting `x^2=y`, we see that, the Integrand is,

`y/(y^2+y-2)=y/((y+2)(y-1))=A/(y+2)+B/(y-1)," say, then"`,

by Heaviside's Method,

`A=[y/(y-1)]_(y=-2)=(-2)/(-2-1)=2/3, and,`

`B=[y/(y+2)]_(y=1)=1/(1+2)=1/3`.

`:. y/(y^2+y-2)=y/((y+2)(y-1))=(2/3)/(y+2)+(1/3)/(y-1)`.

Replacing `y" by "x^2`, we have,

`x^2/(x^4+x^2-2)=(2/3)/(x^2+2)+(1/3)/(x^2-1)`.

`:. intx^2/(x^4+x^2-2)dx=2/3int1/(x^2+2)+1/3int1/(x^2-1)dx`,

`=2/3*1/sqrt2arc tan(x/sqrt2)+1/3*1/(2*1)ln|(x-1)/(x+1)|`,

`=sqrt2/3arc tan(x/sqrt2)+1/6ln|(x-1)/(x+1)|+C,` as respected

HSBC244 has already derived!.