Can you solve this basic kinematics problem?

It is well known that the distance traveled between two points from time ( $a$ to time $b$ ) can be determined by: $x=\int_{a}^{b}{v(t)dt}$ In this problem those two points are $0$ and $\infty$ , so we can write that as: $L=\int_{0}^{\infty}{v(t)dt}$ With the physics of this problem now out of the way, all we have to do now is evaluate this integral.

$\displaystyle L=\int_{0}^{\infty}{\frac { ln(1+t) }{ \sqrt { t }(1 +t) }dt}\quad\quad\color{#3D99F6}{x=\sqrt{t}}$

$\displaystyle L=2\int_{0}^{\infty}{\frac { ln(1+x^2) }{(1 +x^2) }dx}$

To evaluate this, we'll use differentiation under the integral sign :

$\displaystyle L(\lambda)=2\int_{0}^{\infty}{\frac { ln(1+\lambda x^2) }{(1 +x^2) }dx}$

$\displaystyle L'(\lambda)=2\int_{0}^{\infty}{\frac { x^2 }{(1+\lambda x^2)(1 +x^2) }dx}=2\int_{0}^{\infty}{\frac { A}{1+\lambda x^2 }+\frac {B}{1 +x^2 }}$

We can find $A$ and $B$ by using $\frac{ A}{1+\lambda x^2 }+\frac{B}{1 +x^2 }=\frac { x^2 }{(1+\lambda x^2)(1 +x^2) }$ , by solving we find $B=-A=\frac{1}{\lambda-1}$

$\displaystyle L'(\lambda)=\frac{2}{\lambda-1}\int_{0}^{\infty}{\frac { 1}{1 +x^2}-\frac {1}{1 +\lambda x^2 }}$

Using $\displaystyle\int_{0}^{\infty}{\frac{1}{1+x^2}}=\frac{\pi}{2}$ (antiderivative of this function is $\arctan(x)$ ).

$\displaystyle L'(\lambda)=\frac{2}{\lambda-1}(\frac{\pi}{2}-\frac{\pi}{2\sqrt{\lambda}})=\frac{\pi}{\sqrt{\lambda}+\lambda}$

Now we return to $L(\lambda)$ by integrating over $\lambda$ :

$\displaystyle L(\lambda)=\int{\frac{\pi}{\sqrt{\lambda}+\lambda}d\lambda}=2\pi\ln(1+\sqrt{\lambda})+C$

It's obvious that $L(0)=0$ , from this we can find $C=0$ .

$\displaystyle L(\lambda)=2\pi\ln(1+\sqrt{\lambda})$

In our case $\lambda=1$ and from this we find:

$\displaystyle L=2\pi\ln2$

If we define $F(\alpha) \; = \; \int_0^\infty \frac{dt}{\sqrt{t}(1 + t)^\alpha} \; = \; B(\tfrac12,\alpha-\tfrac12)\;, \qquad \alpha > \tfrac12 \;,$ then the required distance is $s \; =\; \int_0^\infty \frac{\ln(1+t)}{\sqrt{t}(1+t)}\,dt \; = \; -F'(1) \;.$ Since $F(\alpha) \; = \; \frac{\Gamma(\frac12)\Gamma(\alpha-\frac12)}{\Gamma(\alpha)} \;,$ we deduce that $F'(\alpha) \,=\, F(\alpha)[\psi(\alpha-\tfrac12) - \psi(\alpha)]$ , and so $s \; = \; F(1)[\psi(1) - \psi(\tfrac12)] \; = \; \pi[-\gamma - (-\gamma-\ln4)] \; = \; \pi\ln 4 \;.$

All we need to do is find the value of the integral $\large \int_{0}^{\infty} \frac{\ln(1+t)}{\sqrt{t} +t\sqrt{t}} dt$

Substituting $\sqrt{t}=z$

We have $\large 2\int_{0}^{\infty} \frac{\ln(1+z^2)}{1 +z^2}dz$

Again substituting $z=tan(y)$

We have

$\large 2\int_{0}^{\frac{\pi}{2}}\ln(1+tan^{2}(y))dy$

$=\large 2\int_{0}^{\frac{\pi}{2}}\ln(sec^{2}(y))dy$

$=\large -4\int_{0}^{\frac{\pi}{2}}\ln(cos(y))dy$

Now this is a well known integral

Here's a youtube video showing how to evaluate this .

And the value of the integral $\int_{0}^{\frac{\pi}{2}}\ln(cos(y))dy = -\frac{\pi}{2}\ln(2)$

So we have our answer is :- $2\pi\ln(2)$