# What is the integral of the equation (3x^2-x+6)/(x-2)(x^2+4)?

The integral is `2 ln | x - 2 | + 1 / 2 ln ( x^2 + 4 ) + 1 / 2 arctan ( x / 2 ) + C .`

To find the indefinite integral of a rational function, one should do partial fraction decomposition. In the case of denominator `( x - 2 ) ( x^2 + 4 ) ` the partial fractions will be `A / ( x - 2 ) ` and `( Bx + C ) / ( x^2 + 4 ) ` for some specific constants `A , ` `B , ` and `C .`

To find them, write the equation and multiply by the denominator:

`( 3x^2 - x + 6 ) / ( ( x - 2 ) ( x^2 + 4 ) ) = A / ( x - 2 ) + ( Bx + C ) / ( x^2 + 4 ) ,`

`3x^2 - x + 6 = A ( x^2 + 4 ) + ( Bx + C ) ( x -2 ) .`

This will lead to `3x^2 - x + 6 = x^2 ( A + B ) + x ( -2B + C ) + ( 4A - 2C ) .`

Because it must be true for any `x , ` we obtain the system of equations

`3 = A + B , ` `-1 = -2B + C , ` `6 = 4A - 2C , ` so `3 = 2A - C ,`

which can be solved this way: `C = 2A - 3 , ` `B = 3 - A , ` `-1 = -6 + 2A + 2A - 3 , ` `4A = 8 , ` `A = 2 . ` Thus, `C = 1 ` and `B = 1 .`

Now it is simple to integrate `2 / ( x - 2 ) + x / ( x^2 + 4 ) + 1 / ( x^2 + 4 ) : ` the first summand gives `2 ln | x - 2 | , ` the second `1 / 2 ln ( x^2 + 4 ) , ` and the third `1 / 2 arctan ( x / 2 ) ` (+ C, of course).

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