Express the integrand as a sum of partial fractions and evaluate the integ $$ \int \frac{d x}{4-25 x^{2}} $$ Express the integrand as a sum of partial fractions.

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Evaluate the integral by following steps:
- step0: Evaluate using substitution:
  $$\int \frac{1}{4-25x^{2}} dx$$
- step1: Transform the expression:
  $$\int \frac{1}{4-4\sin^{2}\left(t\right)}\times \frac{2}{5}\cos\left(t\right) dt$$
- step2: Simplify the expression:
  $$\int \frac{\cos\left(t\right)}{10-10\sin^{2}\left(t\right)} dt$$
- step3: Simplify the expression:
  $$\int \frac{1}{10}\sec\left(t\right) dt$$
- step4: Use properties of integrals:
  $$\frac{1}{10}\times \int \sec\left(t\right) dt$$
- step5: Evaluate the integral:
  $$\frac{1}{10}\ln{\left(\left|\sec\left(t\right)+\tan\left(t\right)\right|\right)}$$
- step6: Substitute back:
  $$\frac{1}{10}\ln{\left(\left|\sec\left(\arcsin\left(\frac{5}{2}x\right)\right)+\tan\left(\arcsin\left(\frac{5}{2}x\right)\right)\right|\right)}$$
- step7: Simplify:
  $$\frac{1}{10}\ln{\left(\left|\frac{1}{\left(1-\frac{25}{4}x^{2}\right)^{\frac{1}{2}}}+\frac{5x}{\left(4-25x^{2}\right)^{\frac{1}{2}}}\right|\right)}$$
- step8: Calculate the absolute value:
  $$\frac{1}{10}\ln{\left(\left|\frac{2}{\left(4-25x^{2}\right)^{\frac{1}{2}}}+\frac{5x}{\left(4-25x^{2}\right)^{\frac{1}{2}}}\right|\right)}$$
- step9: Simplify:
  $$\frac{1}{10}\ln{\left(\left|\frac{2\sqrt{4-25x^{2}}}{\left|4-25x^{2}\right|}+\frac{5x\sqrt{4-25x^{2}}}{\left|4-25x^{2}\right|}\right|\right)}$$
- step10: Add the constant of integral C:
  $$\frac{1}{10}\ln{\left(\left|\frac{2\sqrt{4-25x^{2}}}{\left|4-25x^{2}\right|}+\frac{5x\sqrt{4-25x^{2}}}{\left|4-25x^{2}\right|}\right|\right)} + C, C \in \mathbb{R}$$
To express the integrand $$\frac{1}{4 - 25x^2}$$ as a sum of partial fractions, we first recognize that the denominator can be factored.

The expression $$4 - 25x^2$$ can be rewritten as:

$$
4 - 25x^2 = (2 - 5x)(2 + 5x)
$$

Now, we can express $$\frac{1}{4 - 25x^2}$$ as a sum of partial fractions:

$$
\frac{1}{4 - 25x^2} = \frac{A}{2 - 5x} + \frac{B}{2 + 5x}
$$

To find the constants $$A$$ and $$B$$, we multiply both sides by the denominator $$4 - 25x^2$$:

$$
1 = A(2 + 5x) + B(2 - 5x)
$$

Expanding the right side gives:

$$
1 = (2A + 2B) + (5A - 5B)x
$$

Now, we can set up a system of equations by equating coefficients:

1. For the constant term: $$2A + 2B = 1$$
2. For the coefficient of $$x$$: $$5A - 5B = 0$$

From the second equation, we can simplify:

$$
A = B
$$

Substituting $$A = B$$ into the first equation:

$$
2A + 2A = 1 \implies 4A = 1 \implies A = \frac{1}{4}
$$

Thus, $$B = \frac{1}{4}$$ as well. Therefore, we can express the integrand as:

$$
\frac{1}{4 - 25x^2} = \frac{1/4}{2 - 5x} + \frac{1/4}{2 + 5x}
$$

Now, we can evaluate the integral:

$$
\int \frac{d x}{4 - 25x^2} = \int \left( \frac{1/4}{2 - 5x} + \frac{1/4}{2 + 5x} \right) dx
$$

This can be integrated separately:

$$
= \frac{1}{4} \int \frac{1}{2 - 5x} dx + \frac{1}{4} \int \frac{1}{2 + 5x} dx
$$

The integrals can be evaluated as follows:

1. For $$\int \frac{1}{2 - 5x} dx$$, we use the substitution $$u = 2 - 5x$$, giving $$du = -5dx$$ or $$dx = -\frac{1}{5}du$$:

$$
\int \frac{1}{2 - 5x} dx = -\frac{1}{5} \ln |2 - 5x| + C
$$

2. For $$\int \frac{1}{2 + 5x} dx$$, we use the substitution $$v = 2 + 5x$$, giving $$dv = 5dx$$ or $$dx = \frac{1}{5}dv$$:

$$
\int \frac{1}{2 + 5x} dx = \frac{1}{5} \ln |2 + 5x| + C
$$

Combining these results, we have:

$$
\int \frac{d x}{4 - 25x^2} = \frac{1}{4} \left( -\frac{1}{5} \ln |2 - 5x| + \frac{1}{5} \ln |2 + 5x| \right) + C
$$

Simplifying gives:

$$
= \frac{1}{20} \ln \left| \frac{2 + 5x}{2 - 5x} \right| + C
$$

Thus, the final result of the integral is:

$$
\int \frac{d x}{4 - 25x^2} = \frac{1}{20} \ln \left| \frac{2 + 5x}{2 - 5x} \right| + C
$$
$$ \int \frac{dx}{4 - 25x^2} = \frac{1}{20} \ln \left| \frac{2 + 5x}{2 - 5x} \right| + C $$

Express the integrand as a sum of partial fractions and evaluate the integ \( \int \frac{d x}{4-25 x^{2}} \) Express the integrand as a sum of partial fractions.

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