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Evaluate the following integral 1 / ((x+1)(x+2)2(x+3)3)

by RoRi

Compute the following integral.

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

First, we use partial fraction decomposition. We write

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

This gives us the equation

    \begin{align*} A&(x+2)^2 (x+3)^3 + B(x+1)(x+2)(x+3)^3 + C(x+1)(x+3)^3 \\&+ D(x+1)(x+2)^2(x+3)^2 + E(x+1)(x+2)^2(x+3) + F(x+1)(x+2)^2 = 1. \end{align*}

First, we substitute the values x = -1, x = -2, and x = -3 which gives us

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

Substituting these values of A, C, and F into our equation we have

    \begin{align*} \frac{1}{8}&(x+2)^2(x+3)^3 + B(x+1)(x+2)(x+3)^3 - (x+1)(x+3)^3 \\ &+ D(x+1)(x+2)^2(x+3)^2 + E(x+1)(x+2)^2(x+3) - \frac{1}{2}(x+1)(x+2)^2 = 1. \end{align*}

Now, we substitute the values x = 0, x = 1 and x = 2 to obtain the equations

    \begin{align*} \frac{27}{4} + 54B - 27 + 36D + 12E - 2 &= 1 \\ 72 + 384B - 128 + 288D + 72E - 9 &= 1 \\ 250 + 1500B - 375 + 1200D + 240E - 24 &= 1. \end{align*}

Solving this system we obtain the values

    \[ B = 2, \qquad D = -\frac{17}{8}, \qquad E = -\frac{5}{4}. \]

Therefore, we have the following,

    \begin{align*} \int \frac{dx}{(x+1)(x+2)^2(x+3)^3} &= \int \left( \frac{A}{x+1} + \frac{B}{x+2} + \frac{C}{(x+2)^2} \right.\\ & \left. \qquad + \frac{D}{x+3} + \frac{E}{(x+3)^2} + \frac{F}{(x+3)^3} \right) \, dx \\[10pt] &= \frac{1}{8} \int \frac{dx}{x+1} + 2 \int \frac{dx}{x+2} - \int \frac{dx}{(x+2)^2} \\ & \qquad -\frac{17}{8} \int \frac{dx}{x+3} - \frac{5}{4} \int \frac{dx}{(x+3)^2} - \frac{1}{2} \int \frac{dx}{(x+3)^3} \\[10pt] &= \frac{1}{8}\log |x+1| + 2 \log |x+2| + \frac{1}{x+2} - \frac{17}{8} \log |x+3| \\ & \qquad + \frac{5}{4} \frac{1}{x+3} + \frac{1}{4} \frac{1}{(x+3)^2} + C \\[10pt] &= \frac{4(x+3)^2 + 5(x+2)(x+3) + x + 2 }{4 (x+2)(x+3)^2} \\ & \qquad + \frac{1}{8} \Big( \log|x+1| + 16 \log|x+2| - 17 \log|x+3| \Big) + C\\[10pt] &= \frac{9x^2 + 50x + 68}{4(x+2)(x+3)^2} + \frac{1}{8} \log \left| \frac{(x+1)(x+2)^{16}}{(x+3)^{17}} \right| + C. \end{align*}

One comment

  1. Edwin says:

    Hey! Me again, I just wanted to point out that you have a typo in the first equation when substituting x = 0, the first coefficient should be \frac{27}{2} and not \frac{27}{4} . As always, thank you very much for the solutions!

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