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Evaluate the integral of (2 – x – x2)1/2 / x2

by RoRi

Compute the following integral.

    \[ \int \frac{\sqrt{2 - x - x^2}}{x^2} \, dx. \]

Following Apostol’s hint we multiply the numerator and denominator by \sqrt{2 - x - x^2},

    \begin{align*} \int \frac{\sqrt{2-x-x^2}}{x^2} \, dx &= \int \frac{2-x-x^2}{x^2 \sqrt{2-x-x^2}} \, dx \\[9pt] &= \int \frac{4-x \, dx}{x^2 \sqrt{2-x-x^2}} - \int \frac{2 \, dx}{x^2 \sqrt{2-x-x^2}} - \int \frac{dx}{\sqrt{2-x-x^2}}. \end{align*}

Now, we’ll evaluate these three integrals separately. For the first integral, we make the subsitution

    \[ u = \frac{\sqrt{2-x-x^2}}{x}. \]

This gives us

    \begin{align*} du &= \frac{\left( \frac{-1-2x}{2 \sqrt{2-x-x^2}} \right) x - \sqrt{2-x-x^2}}{x^2} \, dx \\[9pt] &= \frac{(-1-2x)x - 2(2-x-x^2)}{x^2 \sqrt{2-x-x^2}} \, dx\\[9pt] &= \frac{x-4}{x^2\sqrt{2-x-x^2}} \, dx. \end{align*}

Therefore, we have

    \begin{align*} \int \frac{4-x}{x^2\sqrt{2-x-x^2}} \, dx &= - \int du \\[9pt] &= - u \\[9pt] &= - \frac{\sqrt{2-x-x^2}}{x}. \end{align*}

For the second integral… Incomplete.

For the third and final integral in our expression we have

    \begin{align*} \int \frac{dx}{\sqrt{2-x-x^2}} &= \int \frac{dx}{\sqrt{1 - \left( x + \frac{1}{2} \right)^2 + \frac{5}{4}}} \\[9pt] &= \int \frac{dx}{\sqrt{\frac{9}{4} - \left( \frac{2x+1}{2} \right)^2}} \\[9pt] &= \int \frac{\frac{2}{3} \dx}{\sqrt{1 - \left( \frac{2x+1}{3} \right)^2}}. \end{align*}

Now, we make the substitution u = \frac{2x+1}{3}, du = \frac{2}{3} \, dx. This gives us,

    \begin{align*} \int \frac{dx}{\sqrt{2-x-x^2}} &= \int \frac{ \frac{2}{3} \, dx}{\sqrt{1 - \left( \frac{2x+1}{3} \right)^2}} \\[9pt] &= \int \frac{du}{\sqrt{1-u^2}} \\[9pt] &= \arcsin (u) \\[9pt] &= \arcsin \left( \frac{2x+1}{3} \right). \end{align*}

This is incomplete. Otherwise, I’ll try to get back to it as soon as I figure out how to evaluate the second integral above…

6 comments

  1. tom says:

    I got closer using a simpler method, but not 100% sure.Do what Apostol suggested, then, bring a x^2 into the denominator of the square root, making the integral (2-x-x^2)/(x^3*sqrt(2(x^2)-1/x-1)). The derivative under the sqrt is -4/x^3+1/x^2. This we have but two more integrals are created- one resolves to the arcsin portion and the other ends up being integral of 1/(x*sqrt(ax^2+bx+c)). It seems this one is solvable by parts but this is where I’m stuck. It is of the correct form in the integral tables.

  2. tom says:

    I got very close. Multiply the first integral by 1/2, the second one then adds 1/2x instead of 2. The second integral now has an x instead of x^2 in the denominator. Complete the square, substitute with u, then substitute sin(v) for u. Integrand becomes 1/(3cosv-1). Then use tan half angle substitution which results in a messy (for me) log function. My result was quite close but ends up with the 1/2 factor on the first integral and the log was different but I probably just didn’t reduce right.

    • RoRi says:

      That’s an interesting idea. I worked on a bit, but can’t get it to reduce correctly either. I’ll work on it some more later… It seems like it should work, but I don’t know how we’ll ever get rid of that extra factor of 1/2 in the first integral.

      • tom says:

        Yeah, using the 3 integrals but with the 1/2 factor works seems to logical until the end: the middle log term is different then Apostol’s and the arcsin factor gets the 1/2 factor. I did notice someone on math.stackexchange worked this one using an easier route but believe he also had trouble matching the answer.

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