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