The uploaded image contains two integrals that need to be solved:

1. $\int \frac{dx}{\sqrt{25 + 4x^2}}$
2. $\int \frac{1}{x\sqrt{x^4 - 1}} dx$

Let me solve these integrals step by step.

1. Solve $\int \frac{dx}{\sqrt{25 + 4x^2}}$:

We use the substitution method here:

• Let $x = \frac{5}{2} \tan \theta$, so $dx = \frac{5}{2} \sec^2 \theta d\theta$.
• Then, $25 + 4x^2 = 25 + 4\left(\frac{5}{2}\tan\theta\right)^2 = 25\sec^2\theta$.

Substitute: $\int \frac{dx}{\sqrt{25 + 4x^2}} = \int \frac{\frac{5}{2}\sec^2\theta d\theta}{\sqrt{25\sec^2\theta}} = \int \frac{\frac{5}{2}\sec^2\theta}{5\sec\theta} d\theta = \int \frac{1}{2} \sec\theta d\theta$

The integral of $\sec\theta$ is $\ln|\sec\theta + \tan\theta| + C$. Thus: $\int \frac{dx}{\sqrt{25 + 4x^2}} = \frac{1}{2} \ln|\sec\theta + \tan\theta| + C$

Rewriting in terms of $x$:

• Since $x = \frac{5}{2}\tan\theta$, we have $\tan\theta = \frac{2x}{5}$ and $\sec\theta = \sqrt{1 + \tan^2\theta} = \sqrt{1 + \left(\frac{2x}{5}\right)^2} = \frac{\sqrt{25 + 4x^2}}{5}$.

Thus, the solution becomes: $\int \frac{dx}{\sqrt{25 + 4x^2}} = \frac{1}{2} \ln\left|\frac{\sqrt{25 + 4x^2}}{5} + \frac{2x}{5}\right| + C$

2. Solve $\int \frac{1}{x\sqrt{x^4 - 1}} dx$:

Here, we use the substitution $x^2 = \sec u$, so $dx = \frac{\sec u \tan u}{2\sqrt{\sec u}} du = \frac{\sec u \tan u}{2\sqrt{x^2}} du$.

For $x^4 - 1 = (\sec u)^2 - 1 = \tan^2 u$, the integral simplifies after substitution: Let me calculate step by step and simplify furtherimport sympy as sp

Defining variables

x = sp.symbols('x')

First integral

integral_1 = sp.integrate(1 / sp.sqrt(25 + 4 * x**2), x)

Second integral

integral_2 = sp.integrate(1 / (x * sp.sqrt(x**4 - 1)), x)

integral_1, integral_2(asinh(2x/5)/2, Piecewise((Iacosh(x**(-2))/2, 1/Abs(x4) > 1), (-asin(x(-2))/2, True)))(asinh(2x/5)/2, Piecewise((Iacosh(x**(-2))/2, 1/Abs(x4) > 1), (-asin(x(-2))/2, True)))successHere are the results for the given integrals:

1. For $\int \frac{dx}{\sqrt{25 + 4x^2}}$: $\frac{1}{2} \sinh^{-1}\left(\frac{2x}{5}\right) + C$

2. For $\int \frac{1}{x\sqrt{x^4 - 1}} dx$:

   \frac{i}{2} \cosh^{-1}\left(x^{-2}\right) + C & \text{if } \frac{1}{|x^4|} > 1 \\ -\frac{1}{2} \sin^{-1}\left(x^{-2}\right) + C & \text{otherwise} \end{cases}$$

These are the simplified results for the integrals.

Would you like detailed steps for the second integral or clarification on the inverse hyperbolic and trigonometric functions?

Expanding Questions:

1. Can these results be transformed into other forms for specific applications?
2. How do hyperbolic and trigonometric substitutions help in solving integrals?
3. Can we numerically approximate these integrals for specific $x$ values?
4. What are some practical scenarios where such integrals arise?
5. How can substitution choices impact the complexity of solving an integral?

Tip: When facing challenging integrals, look for symmetry, standard forms, or substitution methods that simplify radicals or trigonometric expressions.