How to evaluate Ahmed's integral?

Question

How to show that:

$$\int_{0}^{1}\frac{\tan^{-1}\sqrt{x^{2}+2}}{(x^{2}+1)\sqrt{x^{2}+2}}\mathop{\mathrm{d}x}=\frac{5\pi ^{2}}{96}$$

I saw this on Wolfram.

Answer 1

Let $$I = \int_{0}^{1} \frac{\tan^{-1}(\sqrt{2+x^2})}{(1+x^2)\sqrt{2+x^2}}\text{d}x$$

and $$ f(t) = \int_{0}^{1} \frac{\tan^{-1}(t\sqrt{2+x^2})}{(1+x^2)\sqrt{2+x^2}}dx$$

such that $f(1)$ = $I$.

$$\Rightarrow f'(t) = \int_{0}^{1} \frac{\text{d}x}{(1+x^2)\sqrt{2+x^2}}\frac{\text{d}}{\text{d}t}\tan^{-1}(t\sqrt{2+x^2})$$

$$= \int_{0}^{1} \frac{\text{d}x}{(1+x^2)\sqrt{2+x^2}}\frac{\sqrt{2+x^2}}{{1+ (t\sqrt{2+x^2})^{2}}}$$

$$= \int_{0}^{1} \frac{\text{d}x}{(1+x^2){(1+ 2t^2+t^2x^2)}}$$

Let $$\frac{1}{(1+x^2)(1+ 2t^2+t^2x^2)} \equiv \frac{A}{(1+x^2)} + \frac{B}{(1+ 2t^2+t^2x^2)} $$

So,

$$A(1+ 2t^2+t^2x^2) + {B}{(1+x^2)} \equiv 1 $$

Put $x=0$ to obtain $A(1+2t^2)+B=1$

and $x=1$ to obtain $A(1+3t^2)+2B=1$

On solving those two, we get $$A = \frac{1}{1+t^2}, B = \frac{-t^2}{1+t^2}$$

So, our integral

$$\int_{0}^{1} \frac{\text{d}x}{(1+x^2){(1+ 2t^2+t^2x^2)}} = \int_{0}^{1} {\frac{A}{(1+x^2)} + \int_{0}^{1}\frac{B}{(1+ 2t^2+t^2x^2)}}$$

Putting the values of $A$ and $B$,

$$f'(t) = \frac{1}{1+t^2}\int_{0}^{1} \frac{\text{d}x}{1+x^2} - \frac{t^2}{1+t^2}\int_{0}^{1}\frac{\text{d}x}{1+ 2t^2+t^2x^2}$$

$$\Rightarrow f'(t) = \frac{1}{1+t^2}(\tan^{-1}x)\Biggr|_{0}^{1} - \frac{t}{1+t^2} \int_{x=0}^{x=1}\frac{\text{d}(tx)}{1+ 2t^2+(tx)^2} $$

Let $tx$ = $ u$

$$f'(t) = \frac{1}{1+t^2}(\tan^{-1}(1) - \tan^{-1}(0)) - \frac{t}{1+t^2} \int_{u=0}^{u=t}\frac{\text{d}(u)}{(\sqrt{1+ 2t^2})^{2}+(u)^2} $$

So,

$$f'(t) = \frac{1}{1+t^2}\frac{\pi}4 - \frac{t}{1+t^2} \frac{1}{\sqrt{1+ 2t^2}} \tan^{-1}{\frac{u}{\sqrt{1+ 2t^2}}} \Biggr|_{0}^{t} $$

$$\Rightarrow \frac{\text{d}f}{\text{d}t} = \frac{1}{1+t^2}\frac{\pi}4 - \frac{t}{1+t^2} \frac{1}{\sqrt{1+ 2t^2}} \tan^{-1}{\frac{t}{\sqrt{1+ 2t^2}}}$$

$$\Rightarrow \text{d}f = \frac{\pi}{4}\frac{\text{d}t}{1+t^2}- \frac{t}{1+t^2} \frac{1}{\sqrt{1+ 2t^2}} \tan^{-1}{\frac{t}{\sqrt{1+ 2t^2}}}\text{d}t$$

Integrating both sides from $1$ to $\infty$,

$$\lim_{y \rightarrow \infty} f(y) - f(1) = \frac{\pi}{4}\int_{1}^{\infty}\frac{\text{d}t}{1+t^2}- \int_{1}^{\infty} \frac{t}{1+t^2} \frac{1}{\sqrt{1+ 2t^2}} \tan^{-1}{\frac{t}{\sqrt{1+ 2t^2}}}\text{d}t$$

$$ = \lim_{y \rightarrow \infty} f(y) - I = \frac{\pi}{4}(\tan^{-1}t)\Biggr|_{1}^{\infty} - \int_{1}^{\infty} \frac{t}{1+t^2} \frac{1}{\sqrt{1+ 2t^2}} \tan^{-1}{\frac{t}{\sqrt{1+ 2t^2}}}\text{d}t$$

$$ = \lim_{y \rightarrow \infty} f(y) - I = \frac{\pi}{4}(\frac{\pi}{2}-\frac{\pi}{4}) - \int_{1}^{\infty} \frac{t}{1+t^2} \frac{1}{\sqrt{1+ 2t^2}} \tan^{-1}({\frac{t}{\sqrt{1+ 2t^2}})}\text{d}t$$

$$ = \lim_{y \rightarrow \infty} f(y) - I = \frac{\pi^{2}}{16} - \int_{1}^{\infty} \frac{t}{1+t^2} \frac{1}{\sqrt{1+ 2t^2}} \tan^{-1}({\frac{t}{\sqrt{1+ 2t^2}})}\text{d}t \tag1$$

$$ \lim_{y \rightarrow \infty} f(y) = \lim_{y \rightarrow \infty} \int_{0}^{1} \frac{\tan^{-1}(y\sqrt{2+x^2})}{(1+x^2)\sqrt{2+x^2}}dx$$

$$ = \frac{\pi}{2}\int_{x=0}^{x=1} \frac{1}{(1+x^2)\sqrt{2+x^2}}dx$$

Putting $x=\sqrt{2}\tan \theta$, when $x=0$, $\theta = 0$ and when $x=1$, $\theta = \tan^{-1}(\frac{1}{\sqrt{2}}) $ , also $\text{d}x =\sqrt{2}\sec^2\theta \text{d}\theta $

Substituting, we get

$$\lim_{y \rightarrow \infty} f(y) = \frac{\pi}{2}\int_{0}^{\tan^{-1}(\frac{1}{\sqrt{2}})}\frac{\sqrt{2}\sec^2\theta}{(1+2\tan^2\theta)(\sqrt{2}\sec\theta)}\mathrm d\theta\\ =\int_{0}^{\tan^{-1}(\frac{1}{\sqrt{2}})}\frac{\sec \theta}{1+2\tan^2\theta}\mathrm d\theta $$

$$ = \frac{\pi}{2}\int_{0}^{\tan^{-1}(\frac{1}{\sqrt{2}})}\frac{\frac{1}{\cos\theta}\text{d}\theta}{1+2\frac{\sin^2\theta}{\cos^2\theta}}$$

$$ =\frac{\pi}{2} \int_{0}^{\tan^{-1}(\frac{1}{\sqrt{2}})}\frac{\cos\theta}{\cos^2\theta+2{\sin^2\theta}}\text{d}\theta$$

$$ = \frac{\pi}{2}\int_{\theta = 0}^{\theta = \tan^{-1}(\frac{1}{\sqrt{2}})}\frac{\cos\theta}{1+{\sin^2\theta}}\text{d}\theta$$

Let $\phi = \sin\theta$, so $\text{d}\phi = \cos\theta\text{d}\theta$

When $\theta = 0$, $\phi = 0$, and when $\theta = \tan^{-1}(\frac{1}{\sqrt{2}})$, $\phi = \sin(\tan^{-1}(\frac{1}{\sqrt{2}})) = \frac{1}{\sqrt{3}}$

So,

$$ \lim_{y \rightarrow \infty} f(y) = \frac{\pi}{2}\int_{\phi = 0}^{\phi = \frac{1}{\sqrt{3}}} \frac{\text{d}\phi}{1 + \phi^2}$$

$$= \frac{\pi}{2}tan^{-1}{\phi} \Biggr|_{0}^{\frac{1}{\sqrt{3}}} = \frac{\pi}{2}(\frac{\pi}{6}-0) = \frac{\pi^2}{12} $$

So, $$ \lim_{y \rightarrow \infty} f(y) = \frac{\pi^2}{12} \tag2$$

Putting this back in $(1)$, We get

$$\frac{\pi^2}{12} - I = \frac{\pi^{2}}{16} - \int_{1}^{\infty} \frac{t}{1+t^2} \frac{1}{\sqrt{1+ 2t^2}} \tan^{-1}({\frac{t}{\sqrt{1+ 2t^2}})}\text{d}t$$

$$ \Rightarrow \frac{\pi^2}{12} - I = \frac{\pi^{2}}{16} + \int_{1}^{\infty} \frac{1}{1+(\frac{1}{t})^2} \frac{1}{\sqrt{2+(\frac{1}{t})^2}} \tan^{-1}(\frac{1}{\sqrt{2+(\frac{1}{t})^2}})\frac{-\text{d}t}{t^2}$$

Now let $\frac{1}{t} = z$

$$ \Rightarrow \frac{\pi^2}{12} - I = \frac{\pi^{2}}{16} + \int_{1}^{0} \frac{1}{1+z^2} \frac{1}{\sqrt{2+z^2}} \tan^{-1}(\frac{1}{\sqrt{2+z^2}})\text{d}z$$

$$\Rightarrow \frac{\pi^2}{12} - I = \frac{\pi^{2}}{16} + \int_{1}^{0} \frac{1}{1+z^2} \frac{1}{\sqrt{2+z^2}} (\frac{\pi}{2}-\tan^{-1}{\sqrt{2+z^2}})\text{d}z$$

$$ \Rightarrow \frac{\pi^2}{12} - I = \frac{\pi^{2}}{16} + \frac{\pi}{2}\int_{1}^{0} \frac{\text{d}z}{(1+z^2){\sqrt{2+z^2}}} - \int_{1}^{0} \frac{1}{1+z^2} \frac{1}{\sqrt{2+z^2}}\tan^{-1}{\sqrt{2+z^2}}\text{d}z$$

$$\Rightarrow \frac{\pi^2}{12} - I = \frac{\pi^{2}}{16} -( \frac{\pi}{2}\frac{\pi}{6})+I$$

$$\Rightarrow \frac{\pi^2}{6} -\frac{\pi^{2}}{16} = 2I$$

$$\Rightarrow I = \frac{\pi^2}{96}$$

So, $$\int_{0}^{1} \frac{\tan^{-1}(\sqrt{x^2+2})}{(\sqrt{x^2+2}) (x^2+1)} \ dx = \frac{5 \pi^2}{96}$$

Answer 2

A few ways to evaluate it can be found here

Zafar Ahmed, Knut Dale, George Lamb: Definitely an Integral: 10884. The American Mathematical Monthly 109(7): 670-671 (2002)

http://www.jstor.org/stable/pdfplus/3072448.pdf

Answer 3

Define $f\left(t\right):=\int_{0}^{1}\frac{\arctan\left(t\sqrt{x^{2}+2}\right)}{\left(x^{2}+1\right)\sqrt{x^{2}+2}}dx$ so$$\begin{align}f\left(\infty\right)&=\frac{\pi}{2}\int_{0}^{1}\frac{dx}{\left(x^{2}+1\right)\sqrt{x^{2}+2}}\\&=\frac{\pi}{2}\left[\arctan\frac{x}{\sqrt{x^{2}+2}}\right]_{0}^{1}\\&=\frac{\pi^{2}}{12},\\f^\prime\left(t\right)&=\frac{1}{1+t^{2}}\int_{0}^{1}\left(\frac{1}{x^{2}+1}-\frac{t^{2}}{1+t^{2}\left(x^{2}+2\right)}\right)dx\\&=\frac{1}{1+t^{2}}\left(\frac{\pi}{4}-\frac{t}{\sqrt{1+2t^{2}}}\arctan\frac{t}{\sqrt{1+2t^{2}}}\right).\end{align}$$Substituting $u=t^{-1}$ and using $\arctan\frac{1}{\theta}=\frac{\pi}{2}-\arctan\theta$, Ahmed's integral $A:=f\left(1\right)$ satisfies$$\frac{\pi^{2}}{12}-A=f\left(\infty\right)-f\left(1\right)=\frac{\pi^{2}}{16}-\left(\frac{\pi^{2}}{12}-A\right)\implies A=\frac{5\pi^{2}}{96}.$$

Answer 4

I am not sure how far it is from the actual solution, but here is mine.
Let $I$ be your integral in question.

Consider \begin{align} J= \int_{0}^{1} \int_{0}^{x} \frac{x}{(x^2+1)(x^2+(2+x^2)y^2)} \ dy \ dx. \end{align}Evaluating $J$ in the order presented gives us that

\begin{align} J= \int_{0}^{1} \frac{\tan^{-1}\left(\frac{y\sqrt{x^2+2}}{x}\right)}{(x^2+1) \sqrt{x^2+2}} \Big|_{y=0}^{y=x} \ dx=I. \end{align}

On the other hand, we reverse the order of integration, justified by Fubini's theorem.

\begin{align} J= \int_{0}^{1} \int_{y}^{1} \frac{x}{(x^2+1)(x^2+y^2(x^2+2))} \ dx \ dy. \end{align}

Using partial fractions (or Mathematica), we see that

\begin{align} J=\int_{0}^{1} \frac{\ln \left( \frac{2(y^4+3y^2)}{(y^2+1)(1+3y^2)} \right)}{2(y^2-1)} \ dy, \end{align} which can then be split apart into \begin{align} J=\int_{0}^{1} \frac{\ln \left( \frac{2}{y^2+1} \right)}{2(y^2-1)} \ dy + \int_{0}^{1} \frac{\ln \left( y^2 \right)}{2(y^2-1)} \ dy + \int_{0}^{1} \frac{\ln \left( \frac{y^2+3}{3y^2+1} \right)}{2(y^2-1)} \ dy. \end{align} Denote \begin{align} &J_1= \int_{0}^{1} \frac{\ln \left( \frac{2}{y^2+1} \right)}{2(y^2-1)} \ dy, \\ &J_2=\int_{0}^{1} \frac{\ln \left( y^2 \right)}{2(y^2-1)} \ dy ,\\ &J_3=\int_{0}^{1} \frac{\ln \left( \frac{y^2+3}{3y^2+1} \right)}{2(y^2-1)} \ dy, \end{align} and consider the double integrals \begin{align} &K_1= -\int_{0}^{1} \int_{0}^{1} \frac{x}{(1+x^2y^2)(1+x^2)} \ dx \ dy , \\ &K_2= \int_{0}^{1} \int_{0}^{\infty} \frac{x}{(1+x^2y^2)(1+x^2)} \ dx \ dy , \\ &K_3= \int_{0}^{1} \int_{0}^{1} \frac{3x}{(3+x^2y^2)(3+x^2)} \ dx \ dy , \\ &K_4= \int_{0}^{1} \int_{0}^{1} \frac{3x}{(1+3x^2y^2)(1+3x^2)} \ dx \ dy. \end{align} Evaluating the double integrals as they are with partial fractions(or Mathematica), we find that \begin{align} &J_1=K_1 , \\ &J_2=K_2 , \\ &J_3=K_3-K_4 . \end{align} Now reverse the order of integration for each double integral. Then \begin{align} &K_1= -\int_{0}^{1} \int_{0}^{1} \frac{x}{(1+x^2y^2)(1+x^2)} \ dy \ dx = \int_{0}^{1} \frac{-\tan^{-1}(x)}{1+x^2} \ dx = -\frac{\pi^2}{32} \\ &K_2= \int_{0}^{\infty} \int_{0}^{1} \frac{x}{(1+x^2y^2)(1+x^2)} \ dy \ dx = \int_{0}^{\infty} \frac{\tan^{-1}(x)}{(1+x^2)} \ dx = \frac{\pi^2}{8} \\ &K_3= \int_{0}^{1} \int_{0}^{1} \frac{3x}{(3+x^2y^2)(3+x^2)} \ dy \ dx = \int_{0}^{1} \frac{\tan^{-1}\left(\frac{x}{\sqrt{3}} \right)}{\sqrt{3}(3+x^2)} \ dx = \frac{\pi^2}{72}\\ &K_4= \int_{0}^{1} \int_{0}^{1} \frac{3x}{(1+3x^2y^2)(1+3x^2)} \ dy \ dx = \int_{0}^{1} \frac{\sqrt{3} \tan^{-1}\left(x\sqrt{3} \right)}{(1+3x^2)} \ dx = \frac{\pi^2}{18}, \end{align} which can be confirmed by $u$ substitutions. Putting everything together, we see that \begin{align*} & I= -\frac{\pi^2}{32}+\frac{\pi^2}{8}+\frac{\pi^2}{72}-\frac{\pi^2}{18}\\ &= \frac{5\pi^2}{96}. \end{align*} Hence, \begin{align} \int_{0}^{1} \frac{\tan^{-1}(\sqrt{x^2+2})}{(\sqrt{x^2+2}) (x^2+1)} \ dx = \frac{5 \pi^2}{96}. \end{align}

Remark

$J_2$ is an integral commonly used to solve the Basel Problem. Many Stack Exchange users (including myself) have posted solutions in Different methods to compute $\sum\limits_{k=1}^\infty \frac{1}{k^2}$ . Also you can view Daniele Ritelli's and Luigi Pace's solutions on JSTOR (I believe).

Answer 5

Let’s start our journey with a simple double integral below: $$ \begin{aligned} \frac{\pi^2}{16}&=\int_0^1 \int_0^1 \frac{d x d y}{\left(1+x^2\right)\left(1+y^2\right)} \\&=\int_0^1 \int_0^1\left(\frac{1}{1+x^2}+\frac{1}{1+y^2}\right) \frac{1}{2+x^2+y^2} d x d y \\ & =2 \int_0^1 \int_0^1 \frac{1}{\left(1+x^2\right)\left(2+x^2+y^2\right)} d y d x \\ & =2 \int_0^1 \frac{1}{1+x^2}\left(\int_0^1 \frac{d y}{2+x^2+y^2}\right) d x \\ & =2 \int_0^1 \frac{1}{1+x^2}\left[\frac{1}{\sqrt{2+x^2}} \tan ^{-1} \left(\frac{y}{\sqrt{2+x^2}}\right)\right]_0^1 d x \\ & =2 \int_0^1 \frac{1}{\left(1+x^2\right) \sqrt{2+x^2}} \cdot \tan ^{-1}\left(\frac{1}{\sqrt{2+x^2}}\right) d x \\ & =2 \int_0^1 \frac{\frac{\pi}{2}-\tan ^{-1}\left(\sqrt{2+x^2}\right)}{\left(1+x^2\right) \sqrt{2+x^2}} d x \\ & =\pi \int_0^1 \frac{d x}{\left(1+x^2\right) \sqrt{2+x^2}}-2 \int_0^1 \frac{\tan ^{-1}\left(\sqrt{2+x^2}\right)}{\left(1+x^2\right) \sqrt{2+x^2}} d x \\ & =\frac{\pi^2}{6}-2 \int_0^1 \frac{\tan ^{-1}\left(\sqrt{2+x^2}\right)}{\left(1+x^2\right) \sqrt{2+x^2}} d x \\ & \end{aligned} $$ Re-arranging yields our integral $$ \boxed{\int_0^1 \frac{\tan ^{-1} \left(\sqrt{2+x^2}\right)}{\left(1+x^2\right) \sqrt{2+x^2}} d x=\frac{5 \pi^2}{96}} $$

Answer 6

$$ \text{The general form of Ahmed's integral} $$

$$ I = \int_{0}^{1} \frac{\tan^{-1}(\sqrt{2+x^2})}{(1+x^2)\sqrt{2+x^2}} \, dx $$

To solve this integral, we need to use Feynman's integration technique, also known as Leibniz's formula. We introduce a parameter into the integral, transforming it into the following form: (For a detailed explanation of Leibniz's formula, refer to my fourth article on Gaussian integrals.)

$$ I(u) = \int_{0}^{1} \frac{\tan^{-1}(u\sqrt{2+x^2})}{(1+x^2)\sqrt{2+x^2}} \, dx $$

After introducing the parameter, $I(1)$ corresponds to Ahmed's integral.

Next, let's observe the mathematician's approach and then analyze it.

First, let $u \rightarrow \infty$:

$$ I(\infty) = \frac{\pi}{2} \int_{0}^{1} \frac{1}{(1+x^2)\sqrt{2+x^2}} \, dx $$

Now, we apply a differentiation identity:

$$ \frac{d}{dx}\tan^{-1}[f(x)] = \frac{1}{1+f^2(x)} \left(\frac{df}{dx}\right) $$

Using the observation:

$$ f(x) = \frac{x}{\sqrt{2+x^2}} $$

$$ \frac{d}{dx}\tan^{-1}\left[\frac{x}{\sqrt{2+x^2}}\right] = \frac{1}{(1+x^2)\sqrt{2+x^2}} $$

Now, substitute this into the integral, canceling out integration and differentiation:

$$ I(\infty) = \frac{\pi}{2} \int_{0}^{1} \frac{d}{dx}\tan^{-1}\left[\frac{x}{\sqrt{2+x^2}}\right] \, dx $$

$$ = \frac{\pi}{2} \left[\tan^{-1}\left[\frac{x}{\sqrt{2+x^2}}\right]\right]_{0}^{1} $$

$$ = \frac{\pi^2}{12} $$

Now, let's go back to the initial $I(u)$ and apply Leibniz's formula:

$$ \frac{d}{dx}\left[\tan^{-1}[f(u)]\right] = \frac{1}{1+f^2(u)} \left(\frac{df}{du}\right) $$

For $f(u) = u\sqrt{2+x^2}$:

$$ \frac{dI}{du} = \int_{0}^{1} \frac{1}{(1+x^2)\sqrt{2+x^2}} \frac{\sqrt{2+x^2}}{1+u^2(2+x^2)} \, dx $$

$$ = \int_{0}^{1} \frac{1}{(1+x^2)}\frac{1}{1+u^2(2+x^2)} \, dx $$

$$ = \int_{0}^{1} \frac{1}{(1+2u^2+u^2x^2)(1+x^2)} \, dx $$

Now, perform partial fraction decomposition:

$$ \frac{dI}{du} = \int_{0}^{1} \frac{1}{(1+2u^2+u^2x^2)(1+x^2)} \, dx $$

$$ = \int_0^1 \frac{1}{1+u^2}\left[\frac{1}{1+x^2}-\frac{u^2}{1+2u^2+u^2x^2}\right] \, dx $$

$$ = \frac{1}{1+u^2}\int_0^1\left[\frac{1}{1+x^2}-\frac{u^2}{1+2u^2+u^2x^2}\right] \, dx $$

$$ = \frac{1}{1+u^2}\int_0^1\left[\frac{1}{1+x^2}-\frac{1}{\frac{1+2u^2}{u^2}+x^2}\right] \, dx $$

Now, apply the integral form:

$$ \int\frac{dx}{a^2+x^2} = \frac{1}{a}\tan^{-1}\left(\frac{x}{a}\right) + c $$

Integrate $\frac{dI}{du}$:

$$ \frac{dI}{du} = \frac{1}{1+u^2}\int_0^1\left[\frac{1}{1+x^2}-\frac{1}{\frac{1+2u^2}{u^2}+x^2}\right] \, dx $$

$$ = \left[\frac{1}{1+u^2}\right]\left[\tan^{-1}(x)-\frac{u}{\sqrt{1+2u^2}}\tan^{-1}\left(\frac{xu}{\sqrt{1+2u^2}}\right)\right]_0^1 $$

$$ = \left[\frac{1}{1+u^2}\right]\left[\frac{\pi}{4}-\frac{u}{\sqrt{1+2u^2}}\tan^{-1}\left(\frac{u}{\sqrt{1+2u^2}}\right)\right] $$

Now, integrate the resulting expression over $u$ from 1 to $\infty$:

$$ \int_1^\infty\frac{dI}{du} \, du = I(\infty) - I(1) $$

On the right side, we get:

$$ \int_1^\infty\left[\frac{1}{1+u^2}\left(\frac{\pi}{4}-\frac{u}{\sqrt{1+2u^2}}\tan^{-1}\left(\frac{u}{\sqrt{1+2u^2}}\right)\right)\right] \, du $$

Now, combine terms:

$$ = \int_1^\infty\left[\frac{\pi}{4(1+u^2)}-\frac{1}{1+u^2}\frac{u}{\sqrt{1+2u^2}}\tan^{-1}\left(\frac{u}{\sqrt{1+2u^2}}\right)\right] \, du $$

Certainly, continuing from where we left off:

$$ = \frac{\pi}{4}\int_1^\infty \frac{1}{(1+u^2)} \, du - \int_1^\infty \frac{1}{1+u^2}\frac{u}{\sqrt{1+2u^2}}\tan^{-1}\left(\frac{u}{\sqrt{1+2u^2}}\right) \, du $$

The first integral is straightforward to compute:

$$ = \frac{\pi}{4} \left[\frac{\pi}{2} - \frac{\pi}{4}\right] = \frac{\pi^2}{16} $$

Now, let's evaluate the second integral:

$$ \int_1^\infty \frac{1}{1+u^2}\frac{u}{\sqrt{1+2u^2}}\tan^{-1}\left(\frac{u}{\sqrt{1+2u^2}}\right) \, du $$

Here, we observe a convenient feature – the limits of integration from 1 to $\infty$. This allows us to perform a substitution to change the limits to 0 to 1, making it resemble the original integral $I$:

Let $u = \frac{1}{t}$ with $\frac{du}{dt} = -t^{-2}$:

$$ = \int_0^1 \frac{\tan^{-1}\left(\frac{1}{\sqrt{t^2+2}}\right)}{(1+t^2)\sqrt{t^2+2}} \, dt $$

Now, looking back at the original integral we aimed to solve, denoted as $I$:

$$ I = \int_{0}^{1}\frac{\tan^{-1}(\sqrt{2+x^2})}{(1+x^2)\sqrt{2+x^2}}dx $$

It appears that we have approached a similar form, so we can use the identity $\tan^{-1}(x) + \tan^{-1}\left(\frac{1}{x}\right) = \frac{\pi}{2}$ to simplify the expression:

$$ = \frac{\pi}{2}\int_0^1\frac{dt}{(t^2+1)\sqrt{t^2+2}} - \int_{0}^{1}\frac{\tan^{-1}(\sqrt{2+t^2})}{(1+t^2)\sqrt{2+t^2}} \, dt $$

Now, equating the two expressions (original and modified):

$$ I(\infty) - I(1) = \frac{\pi^2}{16} - \frac{\pi}{2}\int_0^1\frac{dt}{(t^2+1)\sqrt{t^2+2}} + \int_{0}^{1}\frac{\tan^{-1}(\sqrt{2+t^2})}{(1+t^2)\sqrt{2+t^2}} \, dt $$

To simplify, let's perform some algebraic manipulation:

$$ 2I(1) = 2I(\infty) - \frac{\pi^2}{16} $$

$$ I(1) = \frac{\pi^2}{12} - \frac{\pi^2}{32} = \frac{5\pi^2}{96} $$

Answer 7

Observing that $$ \int_0^1 \frac{d y}{y^2+x^2+2}=\left[\tan ^{-1}\frac{y}{\sqrt{x^2+2}}\right]_0^1 =\tan ^{-1} \frac{1}{\sqrt{x^2+2}}=\frac{\pi}{2}-\tan ^{-1} \sqrt{x^2+2}, $$ we have $$ \begin{aligned} \int_0^1 \frac{\tan ^{-1}\left(\sqrt{x^2+2}\right)}{\left(x^2+1\right) \sqrt{x^2+2}} d x& =\frac{\pi}{2} \int_0^1 \frac{1}{\left(x^2+1\right) \sqrt{x^2+2}} d x-\int_0^1 \frac{1}{x^2+1}\left(\int_0^1 \frac{1}{y^2+x^2+2} d y\right) d x \\ & =\frac{\pi^2}{12}-\frac{1}{2} \int_0^1 \int_0^1\left(\frac{1}{x^2+1}+\frac{1}{y^2+1}\right) \frac{1}{y^2+x^2+2} d y d x \textrm{ (By symmetry)}\\ & =\frac{\pi^2}{12}-\frac{1}{2} \int_0^1 \int_0^1 \frac{1}{\left(x^2+1\right)\left(y^2+1\right)} d y dx \\ & =\frac{\pi^2}{12}-\frac{1}{2}\left(\int_0^1 \frac{1}{x^2+1} d x\right)\left(\int_0^1 \frac{1}{y^2+1} d y\right) \\ & =\frac{\pi^2}{12}-\frac{1}{2} \cdot \frac{\pi}{4} \cdot \frac{\pi}{4} \\ & =\frac{5}{96} \pi^2 \end{aligned} $$