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I have got such integral $$\int{\frac{\sqrt{x^2+1}}{x+2}dx}$$ and with Maple I got something like this: $$\int\frac{1}{2} + \frac{1+3u^2+4u^3}{-2u^2+2u^4-8u^3}du$$ And I want to know how to achive this changes.

I tried to use WolframAlpha, but there is scarier solution. This integral was for Gaussian quadrature method, so it's analytic solution is horrible.

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Substituting $x=-8u+2u^2$ could get somewhat close to what you are expecting, but not exactly the expression you have. –  Kirthi Raman May 14 '12 at 17:22

1 Answer 1

up vote 2 down vote accepted
  1. We can use the Euler substitution $t=\sqrt{x^{2}+1}-x$ to obtain a rational fraction in terms of $t$ $$\begin{eqnarray*} I =\int \frac{\sqrt{x^{2}+1}}{x+2}\mathrm{d}x=\frac{1}{2}\int \frac{1+2t^{2}+t^{4}}{t^{2}\left( -1+t^{2}-4t\right) }\mathrm{d}t. \end{eqnarray*}$$
  2. Since the integrand is a rational fraction, we can expand it into partial fractions and integrate each fraction. $$\begin{equation*} \frac{1+2t^{2}+t^{4}}{t^{2}\left( -1+t^{2}-4t\right) }=1-\frac{1}{t^{2}}+ \frac{4}{t}+\frac{20}{t^{2}-4t-1}. \end{equation*}$$

Added. Detailed evaluation. From $t=\sqrt{x^{2}+1}-x$, we get $x=\dfrac{1-t^{2}}{2t}$ and $\dfrac{dx}{dt}=-\dfrac{t^{2}+1}{2t^{2}}$. So we have $$\begin{eqnarray*} I &=&\int \frac{\sqrt{x^{2}+1}}{x+2}\mathrm{d}x=\int \frac{t+\frac{1-t^{2}}{2t}}{\frac{1-t^{2}}{2t}+2}\left( -\frac{t^{2}+1}{2t^{2}}\right) \mathrm{d}t \\ &=&\frac{1}{2}\int \frac{1+2t^{2}+t^{4}}{t^{2}\left( -1+t^{2}-4t\right) } \mathrm{d}t. \end{eqnarray*}$$

Expanding into partial fractions as above, we obtain $$ \begin{eqnarray*} 2I &=&\int 1-\frac{1}{t^{2}}+\frac{4}{t}+\frac{20}{t^{2}-4t-1}\mathrm{d}t \\ &=&\int 1\mathrm{d}t-\int \frac{1}{t^{2}}\mathrm{d}t+4\int \frac{1}{t} \mathrm{d}t+20\int \frac{1}{t^{2}-4t-1}\mathrm{d}t \\ &=&t+\frac{1}{t}+4\ln \left\vert t\right\vert -2\sqrt{5}\ln \frac{\sqrt{5}t-2 \sqrt{5}+5}{5-\sqrt{5}t+2\sqrt{5}}+C \\ &=&\sqrt{x^{2}+1}-x+\frac{1}{\sqrt{x^{2}+1}-x}+4\ln \left( \sqrt{x^{2}+1} -x\right) \\ &&-2\sqrt{5}\ln \frac{\sqrt{5}\left( \sqrt{x^{2}+1}-x\right) -2\sqrt{5}+5}{5-\sqrt{5}\left( \sqrt{x^{2}+1}-x\right) +2\sqrt{5}}+C. \end{eqnarray*}$$

Therefore the given integral is $$\begin{eqnarray*} I &=&\frac{1}{2}\left( \sqrt{x^{2}+1}-x\right) +\frac{1}{2}\frac{1}{\sqrt{ x^{2}+1}-x}+2\ln \left( \sqrt{x^{2}+1}-x\right) \\ &&-\sqrt{5}\ln \frac{\sqrt{5}\left( \sqrt{x^{2}+1}-x\right) -2\sqrt{5}+5}{5-\sqrt{5}\left( \sqrt{x^{2}+1}-x\right) +2\sqrt{5}}+C. \end{eqnarray*}$$

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