Questions: Find the antiderivative for each function when C equals 0. Do as many as you can mentally. Check your answers by differentiation. a. 12/(sqrt(1-x^2)) b. 1/(7(x^2+1)) c. 1/(1+64 x^2)

Solution Steps

To find the antiderivative of each function, we will use known antiderivative formulas and properties of integrals. We will then verify our results by differentiating the antiderivatives.

This function resembles the derivative of the arcsine function.

This function resembles the derivative of the arctangent function.

This function can be transformed to resemble the derivative of the arctangent function by factoring out constants.

The antiderivative of the function \( \frac{12}{\sqrt{1-x^{2}}} \) is given by: \[ \int \frac{12}{\sqrt{1-x^{2}}} \, dx = 12 \arcsin(x) + C \] Setting \( C = 0 \), we have: \[ \int \frac{12}{\sqrt{1-x^{2}}} \, dx = 12 \arcsin(x) \]

The antiderivative of the function \( \frac{1}{7(x^{2}+1)} \) is: \[ \int \frac{1}{7(x^{2}+1)} \, dx = \frac{1}{7} \arctan(x) + C \] Setting \( C = 0 \), we have: \[ \int \frac{1}{7(x^{2}+1)} \, dx = \frac{1}{7} \arctan(x) \]

The antiderivative of the function \( \frac{1}{1+64x^{2}} \) can be expressed as: \[ \int \frac{1}{1+64x^{2}} \, dx = \frac{1}{8} \arctan(8x) + C \] Setting \( C = 0 \), we have: \[ \int \frac{1}{1+64x^{2}} \, dx = \frac{1}{8} \arctan(8x) \]

Final Answer