How to evaluate $\int_{-\infty}^\infty \frac{x-2}{(x^2+x+4)\sqrt{x^2+2x+4}}dx $

Question

I would like to study how to evaluate the definite integral$$\int_{-\infty}^\infty \frac{x-2}{(x^2+x+4)\sqrt{x^2+2x+4}}dx $$

Based on the integration techniques that I know, I proceed by completing the square and then introducing the trigonometric substitution $x+1=\sqrt3 \sinh t$ to transform the integral to $$ \int_{-\infty}^\infty \frac{\sqrt3 \sinh t-3 }{3\sinh^2t-\sqrt3\sinh t +4}dt $$ At this point, I further introduce the ‘half-angle’ substitution $y=\tanh\frac t2$. But, the resulting integrand is of a quartic function in $y$, which I do not how how to deal with. I prefer using real method.

Answer 1

Continue with the trig-substitution \begin{align} & \int_{-\infty}^\infty \frac{\sqrt3 \sinh t-3 }{3\sinh^2t-\sqrt3\sinh t +4}\ dt\\ =& \int_{-\infty}^\infty \frac{(\sqrt3\sinh t-3) \,\text{sech}^2t }{(3\cosh^2t-\sqrt3\sinh t +1)\, \text{sech}^2t}\ dt \\ =& \int_{-\infty}^\infty \frac{\sqrt3 \tanh t \,\text{sech}\,t-3\,\text{sech}^2t}{3 -\sqrt3 \,\text{sech}\,t \tanh t+ \text{sech}^2t}\ dt \\ =& \ {2\sqrt3}\int_{-\infty}^\infty \frac{d(\text{sech}\,t+\sqrt3\tanh t )}{(\text{sech}\,t+\sqrt3\tanh t)^2-9}\\ =& \ \frac2{\sqrt3}\ln(2-\sqrt{3}) \end{align}

Answer 2

$$\begin{align}\int_{-\infty}^\infty\frac{x-2}{(x^2+x+4)\sqrt{x^2+2x+4}}\mathrm dx&\overset{t=\frac{2-x}{2+x}}{=}8\int_{-\infty}^1\frac{t\,\mathrm dt}{(3t^2+5)\sqrt{t^2+3}}\\&\overset{u=\sqrt{t^2+3}}{=}8\int_2^\infty\frac{\mathrm du}{4-3u^2}\\&=\frac{-4}{\sqrt3}\coth^{-1}\sqrt3+C\end{align}$$

Credit:

I got the idea of the substitution $t=\dfrac{2-x}{2+x}$ from @ClaudeLeibovici's answer.

Answer 3

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Let $\mathcal{I}$ denote the value of the following (convergent) improper integral:

$$\mathcal{I}:=\int_{-\infty}^{\infty}\mathrm{d}x\,\frac{x-2}{\left(x^{2}+x+4\right)\sqrt{x^{2}+2x+4}}\approx-1.52069.$$

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Using the Euler substitution

$$\sqrt{x^{2}+2x+4}=x+y\implies x=\frac{4-y^{2}}{2(y-1)},$$

we can rewrite $\mathcal{I}$ as an integral of a rational function:

$$\begin{align} \mathcal{I} &=\int_{-\infty}^{\infty}\mathrm{d}x\,\frac{x-2}{\left(x^{2}+x+4\right)\sqrt{x^{2}+2x+4}}\\ &=\int_{+\infty}^{1}\mathrm{d}y\,\frac{(y^{2}+4y-8)}{2(y-1)^{2}}\cdot\frac{1}{\left[\frac{4-y^{2}}{2(y-1)}\right]^{2}+\left[\frac{4-y^{2}}{2(y-1)}\right]+4};~~~\small{\left[x=\frac{4-y^{2}}{2(y-1)}\right]}\\ &=-\int_{1}^{\infty}\mathrm{d}y\,\frac{2y^{2}+8y-16}{y^{4}-2y^{3}+10y^{2}-24y+24}.\\ \end{align}$$

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We'd like to find a way to express the quartic in the denominator of the integrand as a difference of squares of the form

$$y^{4}-2y^{3}+10y^{2}-24y+24=\left(y^{2}+ay+b\right)^{2}-\left(cy+d\right)^{2},$$

for suitable values of $(a,b,c,d)\in\mathbb{R}^{4}$. Expanding the RHS and matching coefficients, it can be shown that

$$y^{4}-2y^{3}+10y^{2}-24y+24=\left(y^{2}-y+6\right)^{2}-3\left(y+2\right)^{2}.$$

We can then use partial fractions and integrate to obtain

$$\begin{align} \mathcal{I} &=-\int_{1}^{\infty}\mathrm{d}y\,\frac{2y^{2}+8y-16}{y^{4}-2y^{3}+10y^{2}-24y+24}\\ &=-\int_{1}^{\infty}\mathrm{d}y\,\frac{2y^{2}+8y-16}{\left(y^{2}-y+6\right)^{2}-3\left(y+2\right)^{2}}\\ &=-\int_{1}^{\infty}\mathrm{d}y\,\frac{2y^{2}+8y-16}{\left[\left(y^{2}-y+6\right)-\sqrt{3}\left(y+2\right)\right]\left[\left(y^{2}-y+6\right)+\sqrt{3}\left(y+2\right)\right]}\\ &=-\int_{1}^{\infty}\mathrm{d}y\,\frac{2y^{2}+8y-16}{\left[y^{2}-\left(1+\sqrt{3}\right)y+6-2\sqrt{3}\right]\left[y^{2}-\left(1-\sqrt{3}\right)y+6+2\sqrt{3}\right]}\\ &=-\int_{1}^{\infty}\mathrm{d}y\,\frac{1}{\sqrt{3}}\left[\frac{2y-\left(1+\sqrt{3}\right)}{y^{2}-\left(1+\sqrt{3}\right)y+6-2\sqrt{3}}-\frac{2y-\left(1-\sqrt{3}\right)}{y^{2}-\left(1-\sqrt{3}\right)y+6+2\sqrt{3}}\right]\\ &=-\frac{1}{\sqrt{3}}\int_{1}^{\infty}\mathrm{d}y\,\frac{d}{dy}\left[\ln{\left(y^{2}-\left(1+\sqrt{3}\right)y+6-2\sqrt{3}\right)}-\ln{\left(y^{2}-\left(1-\sqrt{3}\right)y+6+2\sqrt{3}\right)}\right]\\ &=\frac{1}{\sqrt{3}}\left[\ln{\left(1-\left(1+\sqrt{3}\right)+6-2\sqrt{3}\right)}-\ln{\left(1-\left(1-\sqrt{3}\right)+6+2\sqrt{3}\right)}\right]\\ &=-\frac{1}{\sqrt{3}}\left[\ln{\left(2+\sqrt{3}\right)}-\ln{\left(2-\sqrt{3}\right)}\right]\\ &=-\frac{2}{\sqrt{3}}\ln{\left(2+\sqrt{3}\right)}.\blacksquare\\ \end{align}$$

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Answer 4

I've sat looking at this for hours, until I found this out. During half angle substituting, I'm sure you will get $$ \int_{-∞}^{∞} \frac{\frac{2\sqrt 3 \tanh (\frac{t}{2})}{1-\tanh^2(\frac{t}{2})} -3}{-\frac{2\sqrt 3 \tanh (\frac{t}{2})}{1-\tanh^2(\frac{t}{2})}+12\big(\frac{ \tanh (\frac{t}{2})}{1-\tanh^2(\frac{t}{2})}\big)^2+4}dt$$ Or $$\int_{-∞}^{∞} \frac{2\sqrt 3 \tanh (\frac{t}{2})(1-\tanh^2(\frac{t}{2}))-3(1-\tanh^2(\frac{t}{2}))^2}{(-2\sqrt 3 \tanh (\frac{t}{2}))(1-\tanh^2(\frac{t}{2}))+12\big({ \tanh (\frac{t}{2})}\big)^2+4(1-\tanh^2(\frac{t}{2}))^2}dt$$

Taking $k=\tanh\frac{t}{2} \implies dk=\frac{\text{sech} ^2\frac{t}{2}}{2} dt =\frac{1-k^2}{2} dt$

Thus $\int_{-1}^{1} \frac{2\sqrt 3k(1-k^2)-3(1-k^2)^2}{-2\sqrt 3k(1-k^2)+12k^2+4(1-k^2)^2}×\frac{2}{1-k^2} dk$

Or $$\int_{-1}^{1} \frac{2\sqrt 3k -3(1-k^2)}{\sqrt 3k(k^2-1)+6k^2+2(1-k^2)^2} dk$$ Or $$\int_{-1}^{1} \frac{3k^2-2\sqrt 3 k -3}{2k^4+\sqrt 3 k^3 + 2k^2-\sqrt 3k+2} dk$$ We first simplify the denominator. This is the part that took so much time. We can write it as $$2k^4+(2\sqrt 3 k^3 -\sqrt 3 k^3) + (4k^2 -3k^2)+ k^2 +(\sqrt 3 k - 2\sqrt 3 k) +2$$ Or $$2k^2(k^2+\sqrt 3 k +2)+(k^2+\sqrt 3k +2)-\sqrt 3 k^3 -3k^2-2\sqrt 3k$$

Even I don't know how I found this. But this is factorisable!!!

This is $$=(2k^2+1-\sqrt 3k)(k^2+\sqrt 3k+2)$$

Thus our integral becomes $$\int \frac{3k^2-2\sqrt 3k-3}{(2k^2+1-\sqrt 3k)(k^2+\sqrt 3k+2)} dk$$

We shall follow partial fraction method, $\int\frac{(Ak+B)(k^2+\sqrt 3k+2)-(Ck+D)(2k^2+1-\sqrt 3k)}{(2k^2+1-\sqrt 3k)(k^2+\sqrt 3k+2)} dk$

By solving simultaneous equations, we get $$A=\frac{4\sqrt 3}{3},B=-1,C=\frac{2\sqrt 3}{3},D=1$$

Thus, our integral becomes $$\frac{1}{3}\int_{-1}^{1} \frac{4\sqrt 3k -3}{2k^2-\sqrt 3k+1} dk - \frac{1}{3}\int_{-1}^{1} \frac{2\sqrt 3 k +3}{k^2+\sqrt 3 k+2}dk$$ $$=\frac{\sqrt 3}{3}\int_{-1}^{1} \frac{4k-\sqrt 3}{2k^2-\sqrt 3k+1} dk - \frac{\sqrt 3}{3}\int_{-1}^{1} \frac{2k+\sqrt 3 }{k^2+\sqrt 3 k+2}dk$$ $$=\frac{1}{\sqrt 3} \bigg[\ln(|2k^2-\sqrt 3k+1|)\bigg|_{-1}^{1}-\ln(|k^2+\sqrt 3k+2|)\bigg|_{-1}^{1}\bigg]$$ $$=\frac{1}{\sqrt 3}×2\ln\bigg|\frac{3-\sqrt 3}{3+\sqrt 3}\bigg|$$ $$=\ln\bigg(\bigg|\frac{3-\sqrt 3}{3+\sqrt 3}\bigg|^{\frac{2}{\sqrt 3}}\bigg)\\≈-1.520691992601$$

Answer 5

A long way for an interesting integral.

$$I=\int \frac{x-2}{(x^2+x+4)\sqrt{x^2+2x+4}}\,dx$$

If you start, as @MathArt did, with

$$u=\frac {2-x}{2+x}\quad \implies \quad I=\int\frac{4 u}{ \left(3 u^2+5\right)\sqrt{u^2+3}}\,du$$

$$\frac{4 u}{\left(3 u^2+5\right)}=\frac{2}{3 u+i \sqrt{15}}+\frac{2}{3 u-i \sqrt{15}}$$

So, two antiderivatives to be computed lookonk like $$J=\int \frac{du}{ (3 u+a)\sqrt{u^2+3}}$$

$$u=\sqrt{3} \tan (v)\quad \implies \quad J=\int \frac{dv}{a \cos (v)+3 \sqrt{3} \sin (v)}$$ Using the tangent half-angle substitution $$v=2 \tan ^{-1}(w)\quad \implies \quad J=\int \frac{2}{-a w^2+6 \sqrt{3} w+a}\,dw$$ which is simple $$J=-\frac{2 }{\sqrt{-a^2-27}}\tan ^{-1}\left(\frac{a w-3 \sqrt{3}}{\sqrt{-a^2-27}}\right)$$ So, $$I=-\frac{1}{\sqrt{3}} \left(\tanh ^{-1}\left(\frac{3}{2}-\frac{ i \sqrt{5}}{2} w\right)+\tanh^{-1}\left(\frac{3}{2}+\frac{ i \sqrt{5}}{2} w\right)\right)$$ Back to $x$ and trying to simplify a little bit $$I=\frac{2}{\sqrt{3}}\tanh ^{-1}\left(\frac{-2 x^2+\left(2 \sqrt{x^2+2 x+4}-3\right) x+\sqrt{x^2+2 x+4}-10}{\sqrt{3} \left(\sqrt{x^2+2 x+4}-x+2\right)}\right)$$

Then the result for the definite integral $$\large\color{blue}{\frac{1}{\sqrt{3}}\log \left(7-4 \sqrt{3}\right)}$$

Edit

I we were looking at

$$I=\int_{-p}^{+p} \frac{x-2}{(x^2+x+4)\sqrt{x^2+2x+4}}\,dx$$ using the above and expanding as a series for large values of $p$ $$I=\frac{1}{\sqrt{3}}\log \left(7-4 \sqrt{3}\right)+\frac{4}{p^2}-\frac{19}{2 p^4}+\frac{11}{p^6}+O\left(\frac{1}{p^8}\right)$$

Answer 6

$$\mathcal I=\int_{-\infty}^\infty\frac{x-2}{(x^2+x+4)\sqrt{x^2+2x+4}}\mathrm dx=\underbrace{\int_{-\infty}^0\frac{x-2}{(x^2+x+4)\sqrt{x^2+2x+4}}\mathrm dx}_{\mathcal I_1}+\underbrace{\int_0^\infty\frac{x-2}{(x^2+x+4)\sqrt{x^2+2x+4}}\mathrm dx}_{\mathcal I_2}$$

Now,

$$\mathcal I_1\overset{x\to-x}{=}-\int_0^\infty\frac{x+2}{(x^2-x+4)\sqrt{x^2-2x+4}}\mathrm dx$$ $$\mathcal I_2\overset{x\to\frac4x}{=}-\mathcal I_2\implies \mathcal I_2=0$$

Now, using the technique mentioned in this post to evaluate integrals of the type $\displaystyle\int\frac{L}{Q_1\sqrt{Q_2}}\mathrm dx$,

$$\begin{align}\mathcal I&= -\int_0^\infty\frac{x+2}{(x^2-x+4)\sqrt{x^2-2x+4}}\mathrm dx\\&=-2\int_0^\infty\frac{\frac1{2\sqrt x}+\frac1{x\sqrt x}}{\left(x+\frac4x-1\right)\sqrt{x+\frac4x-2}}\mathrm dx\\&\overset{t=\sqrt x-\frac2{\sqrt x}}{=}-4\int_0^\infty\frac{\mathrm dt}{(t^2+3)\sqrt{t^2+2}}\\&\overset{u=\frac1t}{=}-\frac{2\sqrt2}3\int_0^\infty\frac{u\,\mathrm du}{\left(u^2+\frac13\right)\sqrt{u^2+\frac12}}\\&\overset{v=\sqrt{u^2+\frac12}}{=}\frac{2\sqrt2}3\int_\frac1{\sqrt2}^\infty\frac{\mathrm dv}{\frac16-v^2}\\&=-\frac4{\sqrt3}\coth^{-1}\sqrt3\\&=\frac2{\sqrt3}\ln\left(2-\sqrt3\right)\end{align}$$

Answer 7

The denominator of the integrand is of the form $$(Ax^2+2Bx+C)\sqrt{ax^2+2bx+c}$$ where both the quadratic polynomials have complex roots. In such a case one uses the substitution $x=(pt+q)/(t+1)$ where $p, q$ are roots of $$(aB-bA) z^2-(cA-aC) z+(bC-cB) =0$$ (one can prove that the above equation must have real roots). For current problem $$a=A=1,b=1,B=1/2,c=C=4$$ so that $p, q$ are roots of $-z^2/2+2=0$ ie $p=2,q=-2$ and thus we can put $x=2(t-1)/(t+1)$ to get $$x^2+x+4=\frac{10t^2+6} {(t+1)^2}$$ and $$\sqrt{x^2+2x+4} =\frac{\sqrt{12t^2+4} }{\text {sgn}(t+1)\cdot (t+1)}$$ and $$x-2=-\frac{4}{t+1}$$ and $$dx=\frac{4dt}{(t+1)^2}$$ Note further that the substitution is discontinuous at $x=2$ so we need to split the range of integration as $(-\infty, 2),(2,\infty)$ and these map to $(-1,\infty),(-\infty,-1)$ as far as variable $t $ is concerned.

The desired integral is thus $$4\int_{-\infty} ^{-1}\frac{dt}{(5t^2+3)\sqrt{3t^2+1}}-4\int_{-1}^{\infty}\frac{dt}{(5t^2+3)\sqrt{3t^2+1}} $$ We can change the range of first integral to $(1,\infty)$ by replacing $t$ with $-t$ and then the above difference of two integrals equals $$-4\int_{-1}^1\frac{dt}{(5t^2+3)\sqrt{3t^2+1}}=-8\int_0^1\frac{dt}{(5t^2+3)\sqrt{3t^2+1}}$$ Next we put $u=t/\sqrt{3t^2+1}$ and the integral then equals $$-8\int_0^{1/2}\frac{du}{3-4u^2}=-2\int_0^{1/2}\frac{du}{(3/4)-u^2}$$ which easily evaluates to $$-\frac{2}{\sqrt{3}}\log (2+\sqrt{3})$$