# Integrate $\int \arcsin\left(\frac{x+2}{x^2+4x+13}\right)\,\mathrm dx$

$\def\d{\mathrm{d}}$Any hint on how to compute this integration?$$\int \arcsin\left(\frac{x+2}{x^2+4x+13}\right)\,\d x$$

This is what I have done:$$\int \arcsin\left(\frac{x+2}{x^2+4x+13}\right)\,\d x=\int \arcsin\left(\frac{x+2}{(x+2)^2+9}\right)\,\d x.$$ Substitute $x+2=3 \tan\theta$ and $\d x=3\sec^2\theta\,\d\theta$,\begin{align*} &\mathrel{\phantom{=}} \int \arcsin\left(\frac{x+2}{(x+2)^2+9}\right)\,\d x\\ &=3\int \arcsin\left(\frac{3\tan\theta}{9\tan^2\theta+9}\right)\sec^2\theta\,\d\theta\\ &=3\int \arcsin\left(\frac{3\tan\theta}{9\sec^2\theta}\right)\sec^2\theta\,\d\theta\\ &=3\int \arcsin\left(\frac{\sin\theta\cos\theta}{3}\right)\sec^2\theta\,\d\theta\\ &=3\int \arcsin\left(\frac{\sin2\theta}{6}\right)\sec^2\theta\,\d\theta. \end{align*}

What can I do from here?

• your integral leads to a elliptic function Apr 19, 2017 at 12:20
• I've just started integration so I don't really know what that means. Does it mean that it can't be integrated further? Apr 19, 2017 at 12:23
• You can not get an answer in turns of basic functions (polynomials, exponents, logarithms, trigonometric, etc). WolframAlpha gives a messy form of the answer: initial problem Apr 19, 2017 at 12:26

Substituting $$u = x + 2$$, $$du = dx$$ transforms the integral to $$\int \arcsin \frac{u}{u^2 + a^2} \,du ,$$ where in our case $$a = 3$$, and applying integration by parts with $$v = \arcsin \frac{u}{u^2 + a^2}$$, $$dw = dx$$ yields $$\arcsin \left(\frac{u}{u^2 + a^2}\right) + \int \frac{(a^2 - u^2) u \,du}{(u^2 + a^2)\sqrt{(u^2 + a^2)^2 - u^2}} .$$ As some of the comments hint, generically a quartic polynomial in a square root means that elliptic functions will be required to evaluate the integral---indeed, WolframAlpha and Maple return unpleasant antiderivatives in terms of elliptic functions---but the form of the integrand allows us to evaluate this integral in terms of elementary functions. Substituting $$y = u^2 + a^2$$, $$dy = 2 u \,du$$ transforms the integral in the previous display equation to $$\int \frac{(y - 2 a^2) \,dy}{2 y \sqrt{y^2 - y + a^2}}.$$ The integrand is rational in $$y$$ and the square root of a quadratic polynomial in $$y$$, so it can be rationalized, e.g., with the Euler Substitution $$\sqrt{y^2 - y + a^2} = t - y$$, so that $$y = \frac{t^2 - a^2}{2 t - 1}, \qquad dy = \frac{2 (t^2 - t + a^2)}{(2 t - 1)^2} \,dt,$$ which transforms the integral in $$y$$ to $$\int \frac{t^2 - 4 a^2 t + a^2}{(t^2 - a^2) (2 t - 1)} \,dt = \int \left(\frac{1}{2 t - 1} + \frac{2 a^2}{a^2 - t^2}\right) \,dt = \frac{1}{2} \log (2 t - 1) + 2 a \operatorname{artanh} \frac{t}{a} + C .$$ Back-substituting (and setting $$a = 3$$), and combining terms to minimize the mess yields the general antiderivative $$\begin{multline*} (x + 2) \arcsin \frac{x + 2}{x^2 + 4 x + 13} + \frac{1}{2} \operatorname{arsinh} \frac{2 x^2 + 8 x + 25}{\sqrt{35}} \\- 3 \operatorname{artanh} \frac{x^2 + 4 x + 5}{6 \sqrt{x^4 + 8 x^3 + 41 x^2 + 100 x + 165}} + C. \end{multline*}$$ Differentiating with a c.a.s. verifies the solution.

Another potential route: integrate by parts, factorize the subsequent quartic, then complete some squares to uncover and employ a "natural" substitution of $$y = \frac{9-(x+2)^2}{9+(x+2)^2} \implies x = 3 \sqrt{\frac{1-y}{1+y}} - 2 \implies dx = -3 \sqrt{\frac{1+y}{1-y}} \,\frac{dy}{(1+y)^2}$$

Upon simplification we end up with

\begin{align*} & \int \arcsin\left(\frac{x+2}{x^2+4x+13}\right) \, dx \\ &= x \arcsin \left(\frac{x+2}{(x+2)^2+9}\right) + \int \color{red}{\frac{(x+2)^2-9}{(x+2)^2+9}} \frac x{\sqrt{(x^2+3x+11) (x^2+5x+15)}} \, dx \\ &= x \arcsin \left(\frac{x+2}{(x+2)^2+9}\right) + \int \frac{3y}{y+1} \, \frac{dy}{\sqrt{y^2+35}} - \int \frac{2y}{\sqrt{y^2+35}} \, \frac{dy}{\sqrt{1-y^2}} \end{align*}

The remaining integrals can be evaluated as

\begin{align*} \int \frac{3y}{y+1} \, \frac{dy}{\sqrt{y^2+35}} &= 3 \int \frac{dy}{\sqrt{y^2+35}} - \int \frac{dy}{(y+1) \sqrt{y^2+35}} \\ &= 3 \operatorname{arsinh}\left(\frac y{\sqrt{35}}\right) + 2 \int \frac{dz}{\left(z-\sqrt{35}\right)^2-36} \\ &= 3 \operatorname{arsinh}\left(\frac y{\sqrt{35}}\right) - \frac13 \operatorname{artanh}\left(\frac{z-\sqrt{35}}6\right) + C \\[2ex] \int \frac{2y}{\sqrt{y^2+35}} \, \frac{dy}{\sqrt{1-y^2}} &= -\int \frac{2}{\sqrt{36-w^2}} \, dw \\ &= -2\arcsin\frac w6 + C \end{align*}

with the help of respective substitutions of

$$z = \frac{\sqrt{y^2+35}-\sqrt{35}}y = \frac{y}{\sqrt{y^2+35}+\sqrt{35}} \\ \implies y=\frac{2\sqrt{35}\,z}{1-z^2} \implies dy = \frac{2\sqrt{35}\,(1+z^2)}{(1-z^2)^2} \, dz$$

and

$$w=\sqrt{1-y^2} \implies y=\sqrt{1-w^2} \implies dy = -\frac{w}{\sqrt{1-w^2}} \, dw$$