The wave function of a particle is

$\psi \left(x\right)=\left\{\begin{array}{l}\frac{b}{\left(1+x^2\right)}-1mm\le x<0mm\\ c\left(1+x^2\right)0mm\le x\le 1mm\end{array}\right.$

and zero elsewhere

$\psi \left(x\right)=\left\{\begin{array}{l}\frac{b}{\left(1+x^2\right)}-1mm\le x<0mm\\ c\left(1+x^2\right)0mm\le x\le 1mm\end{array}\right.$

and zero elsewhere

$$\begin{gathered} \underset{x\to 0^2}{\lim }\frac{b^2}{{\left(1+x^2\right)}^2}=\underset{x\to 0^2}{\lim }{c}^2{\left(1+x\right)}^4 \\ \frac{b^2}{{\left(1+0\right)}^2}=c^2{\left(1+0\right)}^2 \\ {b}^2=c^2 \end{gathered}$$

$-2mm\le x\le 2mm$

calculate the value of the probability of finding the particle in the interval $0mm\le x\le 1mm$

First of all, we have to calculate the values of b and c For the probability interpretation of $\psi \left(x\right)$to make sense, the wave function must satisfy the condition

$$\begin{gathered} {\int }_{-\infty }^{+\infty }{\left|\psi \left(x\right)\right|}^{2}dx=1 \\ {\int }_{-\infty }^{-1}{\left|\psi \left(x\right)\right|}^{2}dx+{\int }_{-1}^0{\left|\psi \left(x\right)\right|}^{2}dx+{\int }_0^1{\left|\psi \left(x\right)\right|}^{2}dx+{\int }_1^{+\infty }{\left|\psi \left(x\right)\right|}^{2}dx=1 \\ 0+{\int }_{-1}^0\frac{b^2}{{\left(1+x^2\right)}^2}dx+{\int }_0^1{c}^2{\left(1+x\right)}^{4}dx+0=1 \end{gathered}$$

Since b=c then

$$\begin{gathered} {\int }_{-1}^0\frac{b^2}{{\left(1+x^2\right)}^2}dx+b^2{\int }_0^1{\left(1+x\right)}^{4}dx=1 \\ {b}^2{\int }_{-1}^0\frac{1}{\left(1+x^2\right)}dx+b^2{\int }_0^1{\left(1+x\right)}^{4}dx=1 \end{gathered}$$

$$\begin{gathered} b^2{\left[\frac{1}{2}\left(\frac{x}{x^2+1}+{\tan }^{-1}x\right)\right]}_{-1}^0+b^2{\int }_0^1\left(1+4x^3+6x^2+4x+x^4\right)dx=1 \\ \frac{b^2}{2}\left[0+0\left(\frac{-1}{2}-{\tan }^{-1}1\right)\right]+b^2\left[1+1+2+2+\frac{1}{5}-0\right]=1 \\ \frac{b^2}{2}\left[\frac{1}{2}+\frac{\mathrm{\pi }}{4}\right]+b^2\left(6.2\right)=1 \\ \frac{b^2}{2}\left[\frac{1}{2}+\frac{\mathrm{\pi }}{4}\right]+6.2b^2=1 \\ {b}^2\left[6.45+0.3925\right]=1 \\ {b}^2=\frac{1}{6.8425} \\ =0.146 \\ b=0.382 \end{gathered}$$

There fore the values of b and c are 0.382, 0.382

Now, the probability that the particle will be found to the right of the origin is

$$\begin{gathered} p\left(0mm\le x\le 1mm\right)={\int }_0^1{\left|\psi \left(x\right)\right|}^{2}dx \\ ={\int }_0^1{\left[c{\left(1+x\right)}^2\right]}^{2}dx \\ ={\int }_0^{1}c{\left(1+x\right)}^{4}dx \\ =c^2{\int }_0^1\left(1+4x+6x^2+4x^3+x^4\right)dx \\ =\left(0.146\right){\int }_0^1\left(1+4x+6x^2+4x^3+x^4\right)dx \\ =\left(0.146\right){\left[x+\frac{4x^2}{2}+\frac{6x^3}{3}+\frac{4x^4}{4}+\frac{x^5}{5}\right]}_0^1 \\ =\left(0.146\right)\left(6.2\right)x100\% \\ =90.52\% \\ =91\% \end{gathered}$$