Question

Evaluate the following integrals: $$ \int \frac{x^{3} d x}{\left(x^{2}-2 x+2\right)} $$

Short answer

Question: Evaluate the integral $\int \frac{x^3 dx}{x^2 - 2x + 2}$. Answer: $\frac{1}{2}x^2 - 2x + Ax - \frac{4}{a}\ln{|x - a|} + C$, where $A = \frac{2a - 4}{a^2 - 2a + 2}$ and $B = \frac{-4}{-a}$.

Step-by-step solution

Perform polynomial long division

To do this, divide the numerator, \(x^3\), by the denominator, \(x^2 - 2x +2\). $$ \begin{array}{c|cc cc} \multicolumn{2}{r}{x} & -2 \\ \cline{2-5} x^2 - 2x + 2 & x^3 & & & \\ \cline{2-2} \multicolumn{2}{r}{x^3} & -2x^2 &+ 2x & \\ \cline{2-4} \multicolumn{2}{r}{} & +2x^2 & -2x & \\ \multicolumn{2}{r}{} & 2x^2 & -4x & + 4 \\ \cline{3-5} \multicolumn{2}{r}{} & & 2x & -4 \\ \end{array} $$ So, we rewrite the integral as: $$ \int \left(x - 2 + \frac{2x - 4}{x^2 - 2x + 2}\right)dx $$

Integrate the polynomial part

The integral of the polynomial part is straightforward: $$ \int (x - 2) dx = \frac{1}{2}x^2 - 2x + C_1 $$

Decompose the remaining fraction using partial fractions

The fraction we need to decompose is \(\frac{2x - 4}{x^2 - 2x + 2}\). Let's rewrite it as: $$ \frac{2x - 4}{x^2 - 2x + 2} = \frac{A}{x - a} + \frac{B}{x - b} $$ where \(a\) and \(b\) are the roots of the denominator. The denominator is not factorable, so we will use the following: $$ \frac{2x - 4}{x^2 - 2x + 2} = A + \frac{B(x-a)}{x^2 - 2x + 2} $$

Solve for A and B

To find \(A\) and \(B\), we need to multiply both sides by \(x^2 - 2x + 2\) and solve for \(A\) and \(B\): $$ 2x - 4 = A(x^2 - 2x + 2) + B(x - a) $$ Choose \(x=a\): $$ 2a - 4 = A(a^2 - 2a + 2) $$ Solve for \(A\): $$ A = \frac{2a - 4}{a^2 - 2a + 2} $$ Choose any other value for \(x\) and solve for \(B\): $$ 2x - 4 = \frac{2a - 4}{a^2 - 2a + 2}(x^2 - 2x + 2) + B(x - a) $$ We can take \(x=0\) to keep it simple: $$ B = \frac{-4}{-a} $$ Let's plug back the values of \(A\) and \(B\) into the integral: $$ \int \left(x - 2 + \frac{2a - 4}{a^2 - 2a + 2}\right)dx + \int \left(\frac{-4}{-a(x - a)}\right)dx $$

Integrate the partial fractions

Now integrate the resulting expression: $$ \int (x - 2 + A) dx + \int \left(\frac{-4}{-a(x - a)}\right)dx = \frac{1}{2}x^2 - 2x + Ax + C_1 - \frac{4}{a}\ln{|x - a|} + C_2 $$ Combine the constants \(C_1\) and \(C_2\): $$ \frac{1}{2}x^2 - 2x + Ax - \frac{4}{a}\ln{|x - a|} + C $$ The final answer for the given integral is: $$ \int \frac{x^3 dx}{x^2 - 2x + 2} = \frac{1}{2}x^2 - 2x + Ax - \frac{4}{a}\ln{|x - a|} + C $$