Integral of 1/sqrt(1-x^2) dx

1. The integral of a constant times a function is the constant times the integral of the function:

   $$\int \frac{\left(1 - x^{2}\right)^{2}}{t}\, dx = \frac{\int \left(1 - x^{2}\right)^{2}\, dx}{t}$$

   1. Rewrite the integrand:

      $$\left(1 - x^{2}\right)^{2} = x^{4} - 2 x^{2} + 1$$

   2. Integrate term-by-term:

      1. The integral of $x^{n}$ is $\frac{x^{n + 1}}{n + 1}$ when $n \neq -1$:

         $$\int x^{4}\, dx = \frac{x^{5}}{5}$$

      1. The integral of a constant times a function is the constant times the integral of the function:

         $$\int \left(- 2 x^{2}\right)\, dx = - 2 \int x^{2}\, dx$$

         1. The integral of $x^{n}$ is $\frac{x^{n + 1}}{n + 1}$ when $n \neq -1$:

            $$\int x^{2}\, dx = \frac{x^{3}}{3}$$

         So, the result is: $- \frac{2 x^{3}}{3}$

      1. The integral of a constant is the constant times the variable of integration:

         $$\int 1\, dx = x$$

      The result is: $\frac{x^{5}}{5} - \frac{2 x^{3}}{3} + x$

   So, the result is: $\frac{\frac{x^{5}}{5} - \frac{2 x^{3}}{3} + x}{t}$

2. Now simplify:

   $$\frac{x \left(3 x^{4} - 10 x^{2} + 15\right)}{15 t}$$

3. Add the constant of integration:

   $$\frac{x \left(3 x^{4} - 10 x^{2} + 15\right)}{15 t}+ \mathrm{constant}$$

The answer is:

$$\frac{x \left(3 x^{4} - 10 x^{2} + 15\right)}{15 t}+ \mathrm{constant}$$