# For an integral of a rational function: P(x)/ q(x) dx What conditions determine if we can calculate integral in terms of elementary functions?Consider the general integral of a rational function...

For an integral of a rational function: P(x)/ q(x) dx What conditions determine if we can calculate integral in terms of elementary functions?

Consider the general integral of a rational function (where p and q are polynomial):

P(x)/ q(x) dx

What conditions determine whether or not we can actually calculate this integral in terms of elementary functions? What are the computational hurdles?

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You need to consider the order of polynomial `P(x)` smaller than the order of polynomial `q(x)` , such that `q`` '(x) = Q(x) != P(x) ` and the order of `q(x)` is larger than order of `P(x)` .

If the quotient `(P(x))/(q(x))` follows the conditions above, then you need to use partial fraction decomposition and then you may split the integral, using its property of linearity, into simpler integrals.

The following example will show you how this thing works, such that:

`int 1/((x - 2)(x - 3)) dx`

You need to notice that differentiating the polynomial `((x - 2)(x - 3))` does not give 1, hence, you may use partial fraction decomposition, such that:

`1/((x - 2)(x - 3)) = A/(x - 2) + B/(x - 3` )

Bringing the fractions to a common denominator, yields:

`1 = Ax - 3A + Bx - 2B => 1 = x(A + B) - 3A - 2B`

Equating the coefficients of like powers yields:

`{(A + B = 0),(-3A - 2B = 1):} => 3B - 2B = 1 => B = 1 => A = -1`

`1/((x - 2)(x - 3)) = -1/(x - 2) + 1/(x - 3)`

Integrating both sides yields:

`int 1/((x - 2)(x - 3)) dx = int -1/(x - 2) dx + int 1/(x - 3) dx`

`int 1/((x - 2)(x - 3)) dx = -ln|x - 2| + ln|x - 3| + c`

`int 1/((x - 2)(x - 3)) dx =ln|(x - 3)/(x - 2)| + c`

Hence, the selected example above illustrates how partial fraction decomposition is of big help if substitution does not work.