I faced this strange integral while solving mechanics homework. I couldn't proceed further.

$$\int\frac1{\sqrt{12x + 0.02x^2}}\,\mathrm dx = \frac{2\sqrt x\sqrt{x + 600}\operatorname{arcsinh}\left(0.0408248\sqrt x\right)}{\sqrt{12x + 0.02x^2}} + C$$

I'd like to know how to get this solution; I tried integration by parts and integration by substitution, but all in vain.

  • 2
    $\begingroup$ Try completing the square first. Then a substitution might be easier. $\endgroup$ – user137731 Feb 27 '15 at 21:12
  • $\begingroup$ Did you try completing the square for the expression underneath the square root sign? $\endgroup$ – Raj Feb 27 '15 at 21:12
  • $\begingroup$ I tried everything $\endgroup$ – Ahmed Abd El Mawgood Feb 27 '15 at 21:13
  • 1
    $\begingroup$ Then show us how far you got. Edit your work into the question. $\endgroup$ – user137731 Feb 27 '15 at 21:13
  • $\begingroup$ $u$ substitution. $\int \dfrac{1}{\sqrt{12x+.02x^2}}dx=\dfrac{1}{\sqrt{.02}}\int \dfrac{1}{\sqrt{x}}\dfrac{1}{\sqrt{12/.02+x}}$. Let $a=\sqrt{12/.02}$. Then ignoring the factor of $.02$, this is just $\int\dfrac{1}{\sqrt{x}}\dfrac{1}{\sqrt{a^2+x}}$. Now integrate by substitution. $\endgroup$ – Moya Feb 27 '15 at 21:14

It is often the case that integrals of the form $$\int \frac{dv}{\sqrt{v^2 \pm(\text{stuff})}}$$ can be solved through trigonometric means. Your integral is no different. Notice that $$\begin{align}\frac{1}{\sqrt{12x + 0.02x^2}} = \frac{1}{\sqrt{\frac{1}{50}(x+300)^2-1800}} \\ = \frac{1}{\sqrt{\frac{1}{50}}\sqrt{(x+300)^2-90000}} \\ = \frac{\sqrt{50}}{300\sqrt{\left(\frac{x+300}{300}\right)^2-1}}\end{align}$$ Now if $u = \frac{x+300}{300}$ then $du = \frac{1}{300}dx$. Hence, $$\int\frac1{\sqrt{12x + 0.02x^2}}\,\mathrm dx = \sqrt{50}\int \frac{du}{\sqrt{u^2-1}} $$ You should recognize $\frac{du}{\sqrt{u^2-1}} $ as the derivative of one of the inverse trig functions (albeit a hyperbolic one). Once you know which one, you will have an extremely easy antiderivative to get your answer.


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