We use the property that

• The antii-derivative ofis

to solve the definite integral

And by the corollary of the Fundamental Theorem of Calculus

And since

the definite integral becomes

is an integral of a rational function, so we need to make the denominator into one of the basic rational functions we know the antiderivatives of and then split it up into multiple parts if necessary.

The most obvious thing to do then, is to substitute:

So our new integral is:

which can then be split up into two pieces:

both of these are simple:

And that's the final answer.

The correct answer is.

Instead of evaluating the integral directly using antiderivatives, it is much eaiser to notice that the bounds of integration are negatives of each other, and thatis an odd function

(in other words,)

Hence

. Sinceis odd.

We use integration by parts for this integral.

Let, and, then taking the derivative ofand the antiderivative of, we get, and.

Using the integration by parts formula,, we have

. Start

. Use the integration formula. (do not evaluate the left part yet).

. Evaluate the last integral, and begin to evaluate the entire expression for the antiderivative.

. Evaluate

. Use rules of logs

. Simplify

The correct answer is.

The integral itself is not too difficult to take, simply use the Power Rule on theandterms. The trick is to be careful when integrating.is a constant value (about) not a variable, so it must be integrated accordingly.

This integral requires integration by parts followed by u-substitution. Here are the details

. Start

. Factor out the

Set up integration by parts with,. We then haveand. Afterward, we use the integration by parts formula.

.

Now at this point we use u-substitution to evaluate the 2nd integral. Let, thenand therefore. Substituting into the integral we have

. (Don't forget to change the bounds of integration by plugging them intofor our equation for.)

This integral isn't possible to integrate directly using antiderivatives, but we can still find its value by noticing thatis an odd function, and that our limits of integration are negatives of each other.

Hence

. (Sinceis an odd function)

We can use u-substitution for this integral

. Start

Let, then, and our integral becomes

. (Don't forget to change the bounds of integration by plugging them into ourequation for)

We proceed by using integration by parts.

. Start

Let, then we get
. Then using the integration by parts formula, we get

To evaluate this integral, we must integrate by parts, according to the following formula:

So, we must assign our u and dv, and differentiate and integrate to find du and v, respectively:

The derivative and integral were found using the following rules:

,

Note that we ignore the constant of integration.

Now, use the above formula:

Note that both the product of u and vandthe integral are being evaluated from zero to.

The integral was performed using the following rule:

Simplifying the above results, we get.