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Evaluate the integral of (4x5-1) / (x5+x+1)2

Compute the following integral.

    \[ \int \frac{4x^5 - 1}{(x^5+x+1)^2}. \]


We compute the integral as follows:

    \begin{align*}  \int \frac{4x^5 - 1}{(x^5+x+1)^2} \, dx &= \int \frac{5x^5 + x - x^5 - x -1}{(x^5+x+1)^2} \, dx \\  &= \int \frac{x(5x^4+1) - (x^5+x+1)}{(x^5+x+1)^2} \, dx. \end{align*}

Now we make a substitution, letting

    \[ u =\frac{x}{x^5+x+1} \qquad \implies \qquad du = \frac{x(5x^4+1) - (x^5+x+1)}{(x^5+x+1)^2} \, dx. \]

This gives us,

    \begin{align*}  \int \frac{4x^5 - 1}{(x^5+x+1)^2} \, dx &= \int \frac{x(5x^4+1) - (x^5+x+1)}{(x^5+x+1)^2} \, dx \\  &= - \int du \\  &= -u + C \\  &= \frac{-x}{x^5+x+1} + C. \end{align*}

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