Question

Use partial fractions to find the indefinite integral. $$ \int \frac{x^{2}+12 x+12}{x^{3}-4 x} d x $$

Short answer

The indefinite integral \(\int \frac{x^{2}+12 x+12}{x^{3}-4 x} d x\) is given by \(3\ln{|x|} - 8\ln{|x-2|} + 2\ln{|x+2|} + C\)

Step-by-step solution

Factor the Denominator

We first factor the denominator \(x^{3}-4x = x(x^{2}-4) = x(x-2)(x+2)\)

Partial Fraction Decomposition

We now express the given function as a sum of partial fractions: \[ \frac{x^{2}+12x+12}{x*(x-2)*(x+2)} = \frac{A}{x}+\frac{B}{x-2} + \frac{C}{x+2}\]We multiply through by the common denominator \(x * (x - 2) * (x + 2)\) to clear the fractions:\[ x^{2} + 12x + 12 = A(x - 2)(x + 2) + B * x * (x + 2) + C * x * (x - 2)\]Setting \(x = 0\), \(x = 2\), and \(x = -2\), we can solve for A, B, and C.

Solve for A, B, and C

Setting \(x = 0\), equation becomes \(12 = 4A\), therefore \(A = 3\). Setting \(x = 2\), equation becomes \(32 = 4B\), hence \(B = 8\). Setting \(x = -2\), equation becomes \(-8 = -4C\), thus \(C = 2\). Hence our expression becomes \[ \frac{3}{x} + \frac{8}{x - 2} + \frac{2}{x + 2}\]

Integrate Each Fraction

We now integrate each of these fractions separately. Integration of a constant divided by a variable is very straightforward and simply equals to the constant times the natural logarithm of the absolute value of the variable. \[ \int{{dx * (\frac{3}{x} + \frac{8}{x-2} + \frac{2}{x+2})}} = 3\int{{\frac{dx}{x}}} + 8\int{{\frac{dx}{x - 2}}} + 2\int{{\frac{dx}{x + 2}}} =3\ln{|x|} - 8\ln{|x-2|} + 2\ln{|x+2|} + C\]