Question

Evaluate ∫9x2/(196x2)2 with respect to dx.

A) 9/784ln∣196x2∣+C
B) 9/784 ln ∣196x∣+C
C) 9/784x+C
D) 9/392x+C

Answer 1

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.