Question

integral of x^-4+2x+3​

Answer 1

Answer:

First, we want to use partial fraction decomposition so we write

\[ \frac{1}{x^4-2x^3} = \frac{1}{x^3 (x - 2)} = \frac{A}{x} + \frac{B}{x^2} + \frac{C}{x^3} + \frac{D}{x-2}. \]

This gives us the equation

\[ Ax^2(x-2) + Bx(x-2) + C(x-2) + Dx^3 = 1. \]

Evaluating at x = 0 and x = 2 we obtain the values of C and D,

\begin{align*} -2C &= 1 &\implies \qquad C &= -\frac{1}{2} \\ 8D &= 1 & \implies \qquad D &= \frac{1}{8}. \end{align*}

Using these values of C and D and evaluating at x = 1 and x = -1 we obtain

\begin{align*} -A - B + \frac{1}{2} + \frac{1}{8} &= 1 \\ -3A + 3B + \frac{3}{2} - \frac{1}{8} &= 1. \end{align*}

Solving this system for A and B we obtain

\[ A = -\frac{1}{8}, \qquad B = -\frac{1}{4}. \]

Therefore,

\begin{align*} \int \frac{1}{x^4-2x^3} \, dx &= \int \left( \frac{A}{x} + \frac{B}{x^2} + \frac{C}{x^3} + \frac{D}{x-2} \right) \, dx \\[9pt] &= -\frac{1}{8} \int \frac{1}{x} \, dx - \frac{1}{4} \int \frac{1}{x^2} \, dx -\frac{1}{2} \int \frac{1}{x^3} \,dx + \frac{1}{8} \int \frac{1}{x-2} \, dx \\[9pt] &= -\frac{1}{8} \log |x| + \frac{1}{4x} + \frac{1}{4x^2} + \frac{1}{8} \log |x-2| + C \\[9pt] &= \frac{1}{4x} + \frac{1}{4x^2} + \frac{1}{8} \log \left| \frac{x-2}{x} \right| + C. \end{align*}