Final answer:
The integral of 1/196^2x^2 dx is equal to (1/3)(1/196^2)x^3 + C, where C is the constant of integration.
Step-by-step explanation:
To evaluate the given integral, ∫9x^2/(196x^2)^2 with respect to dx, we need to simplify the expression within the integral. Simplifying the denominator by expanding the power, we get ∫9x^2/196^2x^4 dx.
Next, we can cancel out a common factor of 9 in the numerator and denominator, resulting in ∫x^2/196^2x^4 dx.
Simplifying further, we can combine the x^2 terms by dividing x^2 by x^4, which gives us ∫1/196^2x^2 dx.
Finally, we can integrate this expression by using the power rule of integration. The integral of 1/196^2x^2 dx is equal to (1/3)(1/196^2)x^3 + C, where C is the constant of integration.
Recognizing that the integral of 1/x2 is -1/x, we can then write the integral as (9/196^2) ∫ -dx/x2. Performing the integration, we get -9/196^2 * (1/x) + C, where C is the constant of integration. This simplifies to -9/38416x + C, which is equivalent to 9/38416x + C because constant multiples can be taken out of the integral.