Questions: Express the integrand as a sum of partial fractions [ int fracdx9-49x^2 ]

Solution Steps

To solve the integral \(\int \frac{d x}{9-49 x^{2}}\), we can express the integrand as a sum of partial fractions. The expression \(9 - 49x^2\) can be factored as a difference of squares, which allows us to decompose the fraction into simpler terms that can be integrated individually.

The integrand is given by \(\frac{1}{9 - 49x^2}\). We can factor the denominator as a difference of squares: \[ 9 - 49x^2 = (3 - 7x)(3 + 7x) \]

Express the integrand as a sum of partial fractions: \[ \frac{1}{(3 - 7x)(3 + 7x)} = \frac{A}{3 - 7x} + \frac{B}{3 + 7x} \] Solving for \(A\) and \(B\), we find: \[ A = \frac{1}{6}, \quad B = -\frac{1}{6} \] Thus, the partial fraction decomposition is: \[ \frac{1}{6(3 + 7x)} - \frac{1}{6(3 - 7x)} \]

Integrate each term separately: \[ \int \left( \frac{1}{6(3 + 7x)} - \frac{1}{6(3 - 7x)} \right) \, dx \] This results in: \[ \frac{1}{6} \int \frac{1}{3 + 7x} \, dx - \frac{1}{6} \int \frac{1}{3 - 7x} \, dx \]

The integrals are: \[ \frac{1}{6} \cdot \frac{1}{7} \ln|3 + 7x| - \frac{1}{6} \cdot \frac{1}{7} \ln|3 - 7x| \] Simplifying, we have: \[ \frac{1}{42} \ln|3 + 7x| - \frac{1}{42} \ln|3 - 7x| \]

Final Answer