Question

Integral x by x minus 1 into x minus 2 into x minus 3

Answer 1

Answer:

\[ \int \frac{x \, dx}{(x+1)(x+2)(x+3)}. \]

First, we need to get the partial fraction decomposition of the integrand. To that end write

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

Then we have the equation

\[ A(x+2)(x+3) + B(x+1)(x+3) + C(x+1)(x+2) = x. \]

Plugging in the values x = -1, x = -2, and x = -3 we obtain the following

\begin{align*} 2A &= -1 & \implies \qquad A &= -\frac{1}{2} \\ -B &= -2 & \implies \qquad B &= 2 \\ 2C &= -3 & \implies \qquad C &= -\frac{3}{2}. \end{align*}

Therefore we have

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