When finding a function given the antiderivative, a good approach is to take the derivative of . 

By using chain rule for the ln(4x)term, the derivative 1x is obtained. Power rule is used to differentiate the second term, and the last term, C, is simply a constant.

Therefore, the correct expression is f(x)=1x+3x2.

For this problem, keep in mind that both terms ( and sin) have inner and outer functions. This indicates that chain rule is involved.

Taking the derivative of  results in , so the second term in the expression must change from positive to negative. Lastly, the “C” term must also be included. Therefore, the correct expression for the antiderivative of  is .

The constant of integration, also known as , is used for indefinite integrals (in other words, the set of all possible antiderivatives of a function). This constant is used to communicate that on a connected domain, the indefinite integral is only defined up to an additive constant. 

Essentially, part of the function can be isolated by taking the antiderivative, but the positioning of this function may differ depending on what constants should be present in the equation. 

Because these constants could be either negative or positive, there are no restrictions on the exact sign of . Therefore, the correct answer is “Yes;  allows the expression of a general form of antiderivatives.”

Sometimes when there are multiple terms added or subtracted from one another within the antiderivative, it can be useful to invoke the following rule:

Separating the terms into two smaller antiderivative chunks can help declutter the problem at hand. Rewriting roots to look like exponents can also be useful.

The first chunk of this problem requires power rule, and the second is a trigonometry derivative identity. Finally, both  terms can be combined at the end to create a single constant of integration. Therefore, the correct answer is the following:

This problem looks complicated at first, but it is really just a trigonometry derivative identity with an inner function of . The inner function requires the application of chain rule.

Since , this identity is the starting point for this question.

First, find the general expression of the antiderivative. 

From here, the correct value of  must be identified in order to find the specific function of  from the general expression:

To finish this problem, we substitute  for  in the general expression:

To simplify this problem, it may be useful to split up the integral into more manageable  chunks: 

Another trick is to rewrite roots to look like exponents (in this case,). Then, proceed with taking the antiderivative, paying close attention to power rule:

The  and  terms can then be combined to create the total constant of integration. 

Finally, multiply constants through to identify the correct answer:

Integrals such as this are seen very commonly in introductory calculus courses. It is often useful to look for patterns such as the fact that the polynomial under the radical in our example, , happens to be one order higher than the factor outside the radical,  You know that if you take a derivative of a second order polynomial you will get a first order polynomial, so let's define the variable: 

                                                            (1)

Now differentiate with respect to  to write the differential for , 

                                                            (2)

Looking at equation (2), we can solve for , to obtain  . Now if we look at the original integral we can rewrite in terms of 

Now proceed with the integration with respect to . 

Now write the result in terms of  using equation (1), we conclude,  

Let 

Then 

Now we can substitute

Now we substitute back

We can use substitution for this integral.

Let ,

then .

Multiplying this last equation by , we get .

Now we can make our substitutions

. Start

. Swap out  with , and  with . Make sure you also plug the bounds on the integral into  for  to get the new bounds.

. Factor out the .

. Integrate (absolute value signs are not needed since .)

. Evaluate