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Evaluate the integral of (8x3 + 7) / ((x+1)(2x+1)3)

by RoRi

Compute the following integral.

    \[ \int \frac{8x^3 + 7}{(x+1)(2x+1)^3} \, dx. \]

Since the denominator is already factored into linear terms we can proceed directly with the partial fraction decomposition. We write,

    \[ \frac{8x^3 + 7}{(x+1)(2x+1)^3} = \frac{A}{x+1} + \frac{B}{2x+1} + \frac{C}{(2x+1)^2} + \frac{D}{(2x+1)^3}. \]

This gives us the equation

    \[ A(2x+1)^3 + B(x+1)(2x+1)^2 + C(x+1)(2x+1) + D(x+1) = 8x^3 + 7. \]

First, we can find the values of A and D by evaluating at x = -1 and x = -\frac{1}{2}, respectively. This gives us

    \begin{align*} -A &= -1 & \implies \quad A &= 1 \\ \frac{1}{2}D &= 6 & \implies \quad D &= 12. \end{align*}

Then using these values of A and D we evaluate at x = 0 and x = 1 (these are just convenient values, there isn’t anything special about them) to obtain the two equations

    \begin{align*} 1 + B + C + 12 &= 7 & \implies \quad B + C &= -6 \\ 27 + 18B + 6C + 24 &= 15 & \implies \quad 3B + C &= -6. \end{align*}

Solving these two equations we obtain B = 0 and C = -6. Therefore, we have the following partial fraction decomposition:

    \[ \frac{8x^3 + 7}{(x+1)(2x+1)^3} = \frac{1}{x+1} - \frac{6}{(2x+1)^2} + \frac{12}{(2x+1)^3}. \]

We can now evaluate the integral,

    \begin{align*} \int \frac{8x^3 + 7}{(x+1)(2x+1)^3} \, dx &= \int \frac{dx}{x+1} - 3 \int \frac{2 \, dx}{(2x+1)^2} + 6 \int \frac{2\, dx}{(2x+1)^3} \\ &= \log |x+1| + \frac{3}{2x+1} - \frac{3}{(2x+1)^2} + C. \end{align*}

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