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

by RoRi

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*}

3 comments

  1. Artem says:

    But this does not use any technique of integration – you just found the antiderivative somehow and inserted as the result of integration (since we are integrating a constant 1 after the substitution) :)

    • Anonymous says:

      Trying to factor x^5 +x +1 and then x^3-x^2+1 and doing the system of equations for this annoying integral absolutely sucks

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