Question: $D: \begin{bmatrix}1&4.19\\2&3.40\\3&2.80\\4&2.30\\5&1.99\\6&1.70\\7&1.51\\8&1.34\\9&1.21\\10&1.09\end{bmatrix}$

In this table of data the first column is representing time (in minutes), and the second column is the mass of some element. 

* The element consists of two types of isotopes that decay rapidly.
* The first isotope has a half-life of 3 minutes, and the second of 8 minutes.

How can I figure out the initial mass of the two isotopes using least square method on this table of data?

*I will use Wolfram Mathematica for the calculations*

Answer: 
**Hint 1:** $t \mapsto (1-x)/(1+x)$

> 
>  $$\begin{equation}\displaystyle\int\frac{x-1}{(x+1)\sqrt{x+x^2+x^3}}\,\mathrm{d}x = 2\arccos\left(\frac{\sqrt{x}} {x+1}\right) + \mathcal{C}\end{equation}$$
> 
> 
> 

**Hint 2:** One can show that $t = (1-x)/(1+x)$ is it's own inverse. In other words $x = (1-t)/(1+t)$. Hence the derivative becomes. $\mathrm{d}x = -2t \,\mathrm{d}t/(t+1)^2$.
Similarly we have $\sqrt{x^3+x^2+x} = \sqrt{(t^3+3)(1-t^2)}/(t+1)^2$ so we get a nice cancelation. Explicitly we have
$$
\begin{align\*}
\sqrt{x^3+x^2+x}
& = \sqrt{ \left(\frac{1-t}{1+t}\right)^3 + \left(\frac{1-t}{1+t}\right)^2 + \left(\frac{1-t}{1+t}\right) } \\
& = \sqrt{ \frac{-t^3+t^2-3t+3}{(1+t)^3}} = \sqrt{ \frac{1+t}{1+t}\frac{(1-t)(t^2+3)}{(1+t)^3}} = \frac{\sqrt{(1-t^2)(t^3+3)}}{(t+1)^2}
\end{align\*}
$$

**Hint 3:** Use the substitution $\cos u \mapsto t$, what happens?
$\bf{My\; Solution::}$ Given $$\displaystyle \int \frac{1-x}{(1+x)\sqrt{x+x^2+x^3}}dx = -\int\frac{(x^2-1)}{(x+1)^2\sqrt{x+x^2+x^3}}dx$$

We can Write it as $$\displaystyle -\int\frac{\left(1-\frac{1}{x^2}\right)}{\left(x+\frac{1}{x}+2\right)\sqrt{x+\frac{1}{x}+1}}dx$$

Now Let $$\displaystyle \left(x+\frac{1}{x}+1\right)=t^2\;,$$ Then $$\displaystyle \left(1-\frac{1}{x^2}\right)dx = 2tdt$$

So Integral Convert into $$\displaystyle - \int\frac{2t}{(t^2+1)\cdot t}dt = -\int\frac{2}{1+t^2}dt = -2\tan^{-1}(t)+\mathcal{C}$$

So Integral Convert into $$\displaystyle \int \frac{1-x}{(1+x)\sqrt{x+x^2+x^3}}dx =-2\tan^{-1}\left(\sqrt{x+\frac{1}{x}+1}\right)+\mathcal{C}$$